direct & inverse kinematics1 algorithmic robotics and motion planning (0368.4010.01) instructor:...

Direct & Inverse Kinematics 1

Direct & Inverse Kinematics

Algorithmic Robotics and Motion Planning (0368.4010.01)Instructor: Prof. Dan Halperin


Overview

KinematicsIntroduction to Protein Structure A kinematic View of Loop Closure


Overview

Kinematicsthe science of motion that treats the subject without regard to the forces that cause it

Introduction to Protein StructureA kinematic View of Loop Closure


Direct & inverse kinematics of manipulatorsWhat are we trying to do ? (direct)

Go right !!!

???


Direct & inverse kinematics of manipulatorsWhat are we trying to do ? (inverse)

Take the

ball !!!

???



Spatial description and transformationDirect kinematicsInverse kinematics


Spatial description and transformationWe need to be able to describe the

position and the orientation of the robot’s parts

Suppose there’s a universe coordinate system to which everything can be referenced.


Spatial description and transformationWe need to be able to describe the position and

the orientation of the robot’s parts (relative to U)

Uu n iv e rs e

co o rdin a tes y s te m

What’s its position(“reference point”) ?

What’s its orientation ?



Spatial description and transformation• Spatial description• Transformations• Presentation of orientation

Direct kinematicsInverse kinematics


Positions, orientations and framesThe position of a point p relative to a coordinate

system A (Ap):

x

Ay

z

p

p p

p

ZA

X A

YA

A p


Positions, orientations and framesThe orientation of a body is described by a

coordinate system B attached to the body, relative to A (a known coordinate system).

ZB

X B

ZA

X A

YA

YB


Positions, orientations and framesThe orientation of a body is described by a

coordinate system B attached to the body, relative to A (a known coordinate system).

A A A AB B B B

B A B A B A

B A B A B A

B A B A B A

R X Y Z

X X Y X Z X

X Y Y Y Z Y

X Z Y Z Z Z

cosinus of the angle


Positions, orientations and framesA frame is a set of 4 vectors giving the position

and orientation.Example: frame B

,A AB BorgB R P

ZA

X A

YA

A P B or g

ZB

X B YB


position

A P B or gZA

X A

YA

Positions, orientations and framesRemember the robot’s part:

ZB

X B YB

orientation


Mapping

Until now, we say how to describe positions, orientations and frames.

We need to be able to change descriptions from one frame to another: mapping.

Mappings:– translated frames– rotated frames– general frames


Mappings involving translated framesExpressing a point Bp in terms of frame {A},

when {A} has the same orientation as {B}:A B A

Borgp p p

ZA

X A

YA

A p B or g

ZB

X B

YB

B pA p


Mappings involving rotated framesExpressing a vector Bp in terms of frame {A},

when the origins of frames {A} and {B} are coincident:

ZB

X B

ZA

X A

YA

YB

B p


Ap‘s components are Bp’s projections onto the unit directions of {A}.

Remember the rotation matrix :it’s columns are the unit vectors of {B} expressed in {A}.

Thus:

Mappings involving rotated frames

AB R

A A BBP R P


Mappings involving rotated frames: example Given:frame {B} is rotated relative to frame

{A} about Z by 30 degrees, and BP.Calc: AP

= ZB

X B

ZA X A

YAYB

B p


Mappings involving rotated frames: example Sol:

cos30 cos120 0

cos60 cos30 0

0 0 1

:

!

B A B A B AAB B A B A B A

B A B A B A

A A BB

X X Y X Z X

R X Y Y Y Z Y

X Z Y Z Z Z

And

P R P

exactcomputation

!?


Mappings involving general frames{A} and {B} has different origins and orientations.Vector offset between origins: ApBorg

{B} is rotated in respect to {A}: AB R

ZA

X A

YA

A p B or g

B pA p ZB

X B YB


Mappings involving general framesFirst, describe Bp relative to a frame that has the

same orientation of {A}, but whose origin coincides with the origin of {B}

Then add ApBorg for the translation

Thus:

ZA

X A

YA

A p B or g

B pA p ZB

X B YB

A A B AB BorgP R P P


Mappings involving general frames“Homogeneous transform”:

A “transform” specifies a frame.

1 0 0 0 1 1

A A A BB Borg

A A BB

P R P P

P T P


Multiplication of transforms

Given Cp. We want to find Ap.

ZA

X A

YA

C p

A p ZB

X B YB

ZC

X C

YC


Compound transforms

Given Cp. We want to find Ap.Frame {C} is known relative to frame {B}, and

frame {B} is known relative to frame {A}.

0 0 0 1

B B C A A BC B

A A BC B C

A B A B AB C B Corg Borg

P T P and P T P

T T T

R R R P P


Inverting a transform

Frame {B} is known relative to frame {A}We want the description “frame {A} relative to

frame {B}”Straightforward way: compute the inverse

matrix (of a 4x4 matrix)




frame {B}”Better way:

– Compute

– Compute APBorg:

A AB B( )TR R

BA

B BA A

0 ( )B A A BBorg Borg Aorg

B A T AAorg Borg Borg

P R P P

P R P R P




frame {B}”Better way:

0 0 0 1

A T A T AB B B BorgA

R R PT


Transformations: example


Presentation of Orientation

Rotation matrices are useful as operator.Still, it’s “unnatural” to have to give elements

of a matrix with orthonormal columns as input. There are several presentations which make

that input process easier:– Fixed angles

– Euler angles

– Euler parameters

– Quaternions


Fixed angles

X-Y-Z fixed angles– Start with 2 frames: a fixed reference frame {A} and

a coinciding frame {B}

– First rotate {B} by γ about XA, then by β about YA and finally by α about ZA.

– The equivalent rotation matrix is:0 0 1 0 0

( , , ) 0 0 1 0 0

0 0 1 0 0

AB XYZ

c s c s

R s c c s

s c s c

c c c s s s c c s c s s

s c s s s c c s s c c s

s c s c c


Euler angles

Z-Y-X Euler angles– Start with 2 frames: a moving reference (Euler

angles) frame {B} and a coinciding fixed frame {A}

– First rotate {B} by α about ZB, then by β about YB and finally by γ about XB.

– The equivalent rotation matrix is:

' ' '

0 0 1 0 0

( , , ) 0 0 1 0 0

0 0 1 0 0

AB Z Y X

c s c s

R s c c s

s c s c

c c c s s s c c s c s s

s c s s s c c s s c c s

s c s c c

The final resultis the same asX-Y-Z fixed

angles!!!


Fixed & Euler angles

In general:3 rotations taken about fixed axes (fixed angles) yield the same final orientation as the same 3 rotations taken in opposite order about the axes of the moving frame (Euler angles).

There are other angle-set conventions: Z-Y-Z, etc. (both for fixed and moving reference frames).


Euler parameters

Given an equivalent axis K=[KX KY KZ]T (a unit vector we want to rotate about) and an angle θ, the Euler parameters are defined as:

The rotation matrix Rε is:

1 2 3

4

sin , sin , sin ,2 2 2

cos2

X Y Zk k k

2 22 3 1 2 3 4 1 3 2 4

2 21 2 3 4 1 3 2 3 1 4

2 21 3 2 4 2 3 1 4 1 2

1 2 2 2( ) 2( )

2( ) 1 2 2 2( )

2( ) 2( ) 1 2 2

R


Quaternions

Definition: a generalization of complex numbers, obtained by adding the elements i, j, and k to the real numbers, where i, j, and k satisfy: i2=j2=k2=ijk=-1.

a+bi+cj+dk, with a,b,c and d real numbersQuaternion’s conjugate: a-bi-cj-dkQuaternions are associative, distributive and not

commutative.Another representation: a+vector(b,c,d)


Quaternions

A rotation about the unit vector K=[KX KY KZ]T by an angle θ, can be computed using the quaternion:

p’, the rotation of point p(0,pX,pY,pZ), is given by:

Rotations may be contatenated:

q’s elements are the same as the Euler parameters!

(cos , sin )2 2

q K

��

'p qpq

2 1 1 2 2 1 1 2 2 1 2 1( ) ( ) ( ) ( )q q pq q q q p q q q q pq q



Spatial description and transformationDirect kinematics

– Link description– Link-connection description– Affixing frames to links– Manipulator kinematics– Example

Inverse kinematics


Link Description

Think of the manipulator as a chain of bodies (links) connected by joints.

We will consider manipulators constructed with joints of 1 degree of freedom (DOF): revolute and prismatic joints.

The links are numbered from 0 (immobile base) to n (free end of the arm).


Link Description


Link Description

Joint axis i: the line about which link i rotates relative to link i-1

link i-1 can be specified by 2 numbers: link length ai-1 and link twist αi-1

Link length and twist are sufficient to define the relation between any 2 axes in space





Inverse kinematics


Link-connection description

Neighboring links have a common axis2 parameters define the link-connection:

– Link offset di: the distance along the common axis from one link to the next

– Joint angle θi: amount of rotation about the common axis

The link offset di is variable if joint i is prismatic

The joint angle θi is variable if the joint is revolute



variable offset di

variable angle θi


First and last link in the chain

The link length ai, and the link twist αi depend on the joint axis i and i+1.

Convention:a0=an=0 and α0=αn=0

Similar for the link offset di and the joint angle θi :

if joint 1 is revolute, then d1=0. if joint 1 is prismatic, then θ1=0.


Denavit-Hartenberg notation

Any robot can be described kinematically by 4 quantities for each link:– 2 for the link– 2 to describe the link’s connection

For revolute joints, θi is called the joint variable (the other 3 quantities are fixed).

For prismatic joints, di is the joint variable (the other 3 quantities are fixed).


Denavit-Hartenberg notation

The definition of mechanics by means of these quantities is called the Denavit-Hartenberg notation.





Inverse kinematics


Affixing frames to links

We define a frame attached to each link:frame {i} is attached rigidly to link {i}

Convention:– the origin is located where the “link length

line” ai intersects the joint axis

– the Z axis is coincident with the joint axis

– the X axis points along ai, to the direction from joint i to joint i+1

– The Y axis is formed by the right-hand rule


Affixing frames to links


First and last link in the chain

Frame {0} is the immobile base (link 0) of the robot. Thus a0=0 and α0=0.

If joint 1 is revolute, then d1=0.If joint 1 is prismatic, then θ1=0.

If joint n is revolute, then Xn’s direction is the same as Xn-1’s (θ1=0), and {n}’s origin is the intersection of Xn-1 and axis n when dn=0.


Example

i αi-1 ai-1 di θi

1 0 0 0 θ1

2 0 L1 0 θ2

3 0 L2 0 θ3





Inverse kinematics


Manipulator kinematics

We want to construct the transform that defines frame {i} relative to frame {i-1}, as a function of the four link parameters

Each transform will be a function of only 1 joint variable

Each link has his frame, thus the kinematics problem has been broken into n subproblems 1i

i T



Each transform can be written as a combination of a translation and a rotation

The single transformation that relates frame {n} to frame {0}:0 0 1 1

1 2n

n nT T T T



General form:

1

1 1 1 11

1 1 1 1

cos sin 0

sin cos cos cos sin sin

sin sin cos sin cos cos

0 0 0 1

i i i

i i i i i i iii

i i i i i i i

a

dT

d


Frames with standard names





Inverse kinematics


Example: PUMA 560

PUMA 560 is a 6 DOFs industrial robot with all rotational joints (6R mechanism)


Example: PUMA 560

Frame {0} and Frame {1} coincides when θ1=0.

The joint axes Z4, Z5 and Z6 (wrist’s joints) intersect at a common point.

Z4, Z5 and Z6 are mutually orthogonal.


Example: PUMA 560

Frames and link parameters:


Example: PUMA 560



Example: PUMA 560


αi-1ai-1 di θi

1 0 0 0 θ1

2 -90º 0 0 θ2

3 0 a2 d3 θ3

4 -90º a3 d4 θ4

5 90º 0 0 θ5

6 -90º 0 0 θ6


Example: PUMA 560

Link transformations:1 1 4 4 3

1 1 40 31 4

4 4

2 2 5 5

1 42 5

2 2 5 5

3 3 2

3 323

3

0 0 0

0 0 0 0 1

0 0 1 0 0 0

0 0 0 1 0 0 0 1

0 0 0 0

0 0 1 0 0 0 1 0

0 0 0 0

0 0 0 1 0 0 0 1

0

0 0

0 0 1

0 0

c s c s a

s c dT T

s c

c s c s

T Ts c s c

c s a

s cT

d

6 6

56

6 6

0 0

0 0 1 0

0 0

0 1 0 0 0 1

c s

Ts c


Example: PUMA 560

The kinematics equations of the PUMA 560:

0 0 1 2 3 4 511 1 23 5 6 4 5 23 5 5 1 4 56 1 2 3 4 5 6

11 12 13

21 22 23

31 32 33

1 2 2 3 23 4 23 3 1

1 2 2 3 23 4 23 3 1

3 23 2 2 4 23

[ ( ) ] (

0 0 0 1

:

[ ] ,

[ ] ,

,

X

Y

Z

X

Y

Z

r c c c c s s s s c s s c cT T T T T T T

r r r p

r r r p

r r r p

With

p c a c a c d s d s

p s a c a c d s d c

p a s a s d c

6 4 6

21 1 23 4 5 6 4 6 23 5 6 1 4 5 6 4 6

31 23 4 5 6 4 6 23 5 6

12 1 23 4 5 6 4 6 23 5 6 1 4 6 4 5 6

22 1 23 4 5 6 4 6 23 5 6 1 4 6 4 5 6

32 23 4

),

[ ( ) ] ( ),

( ) ,

[ ( ) ] ( ),

[ ( ) ] ( ),

(

c s

r s c c c c s s s s c c s c c c s

r s c c c s s s s c

r c c c c c s c s s c s c c s c s

r s c c c s s c s s s c c c s c s

r s c c

5 6 4 6 23 5 6

13 1 23 4 5 23 5 1 4 5

23 1 23 4 5 23 5 1 4 5

33 23 4 5 23 5

) ,

( ) ,

( ) ,

.

s s c c s s

r c c c s s c s s s

r s c c s s c c s s

r s c s s c




– About the problem– Method of solution– Example


About the problem

We now want to compute the set of joint variables, given the desired position and orientation of the tool relative to the station.

The problem is non-linear:given , find the values of θ1,…,θn .These equations can be non-linear, transcendental.

0nT


About the problem

Consider the PUMA 560:– Given the matrice , solve PUMA 560’s

kinematics equations for the joint angles θ1 through θ6.

– We get 3 independent equations from the rotation-matrix part of and 3 equations from the position-vector part of .

– Thus: 6 nonlinear, transcendental equations and 6 unknowns, and this for a “very simple” 6 DOFs manipulator!

06T

06T

06T


About the problem

We must ask the following questions:– Is there a solution?– Are there several solutions?– How to solve the problem?


Workspace

If the goal (desired position and orientation) is in the reachable workspace, then there’s at least 1 solution.

The reachable workspace is dependent on the manipulator.

There might be several solutions.We won’t cover the following

considerations:obstacles, limits on joint ranges,...


Multiple solutions


Method of solution

There’s no general algorithmConsider a manipulator as “solvable” if it

is possible to calculate all the solutions.Manipulator solution strategies might be

split into 2 classes: closed-form and numerical solutions (numerical solutions won’t be covered – generally slower because of their iterative nature).


Method of solution

closed-form solutions methods are solutions methods based on analytic expressions or on the solution of a polynomial of degree 4 or less.

There’s a general numerical solution for which all “6 DOFs in a single chain” system with revolute and prismatic joints are solvable!

Only on special cases can they be solved analytically.


Method of solution

There are several closed-form solution strategies. We’ll discuss the following 2:– Algebraic solution by reduction to

polynomial– Pieper’s criteria

We’ll also discuss a heuristically solution strategy:– Cyclic Coordinate Descent (CCD)


Algebraic solution by reduction to polynomialSubstitution:

Advantage:– the substitution yields an expression in terms

of variable ui instead of sin θi and cos θi.

Once the solutions for ui are found, θi=2tan-1(solutions-of-ui).

2

2 2

tan2

1 2cos ,sin

1 1

u

u u

u u


Pieper’s criteria

There’s a closed-form solution for 6 DOFs manipulators (with prismatic and/or revolute joints configurations) in which 3 consecutive axes intersect in 1 point.

Almost every 6 DOFs manipulator built today respect Pieper’s criteria.

For ex: PUMA 560’s joint axes 4, 5 and 6.


Cyclic Coordinate Descent

Algorithm:– adjusting one DOF at a time (iterative) to

minimize tool’s distance to the goal– starts at the last link and works backwards,

adjusting each joint along the way– repeat the whole set until “satisfied” or

maximum nr. of sets reached

Each step results in one equation with one unknown for each degree.



Adjusting one link at the time

Tool’s currentposition

Goal’sposition

minimize

Joint tomove



Adjusting one link at the time



starts at the last link, adjusting each joint along the way

repeat until “satisfied”



Advantages:– Allow constraints to be placed (at each step)– Free of singularities– Degree independent– Computationally inexpensive (fast)– Simple to implement

Disadvantage:– Might not find a solution


Example: PUMA 560

We know the links transformations in θi (1≤i≤6).1 1 4 4 3

1 1 40 31 4

4 4

2 2 5 5

1 42 5

2 2 5 5

3 3 2

3 323

3

0 0 0

0 0 0 0 1

0 0 1 0 0 0

0 0 0 1 0 0 0 1

0 0 0 0

0 0 1 0 0 0 1 0

0 0 0 0

0 0 0 1 0 0 0 1

0

0 0

0 0 1

0 0

c s c s a

s c dT T

s c

c s c s

T Ts c s c

c s a

s cT

d

6 6

56

6 6

0 0

0 0 1 0

0 0

0 1 0 0 0 1

c s

Ts c


Example: PUMA 560

We know the links transformations in θi (1≤i≤6).

Thus, we know the transformation from {k} to {l} in θi (1≤k,l≤6 and k≤i≤l w.l.o.g.).

11

1 1 11

k k ll k l

l l kk l k

T T T and

T T T


Example: PUMA 560

We know the links transformations in θi (1≤i≤6).

Thus, we know the transformation from {k} to {l} in θi (1≤k,l≤6 and k≤i≤l w.l.o.g.).

We wish to solve:11 12 13

21 22 2306

31 32 33

0 1 2 3 4 51 1 2 2 3 3 4 4 5 5 6 6

0 0 0 1

( ) ( ) ( ) ( ) ( ) ( )

X

Y

Z

r r r p

r r r pT

r r r p

T T T T T T


Example: PUMA 560

We’ll search for a solvable equation iteratively:– multiply each side of the transform equation by an

inverse to separate a variable

– Search for a solvable equations0 0 1 2 3 4 56 1 1 2 2 3 3 4 4 5 5 6 6

0 1 0 1 2 3 4 51 1 6 2 2 3 3 4 4 5 5 6 6

1 1 11 12 13

1 1 21 22 23 16

31 32 33

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

0 0

0 0

0 0 1 0

0 0 0 1 0 0 0 1

X

Y

Z

T T T T T T T

T T T T T T T

c s r r r p

s c r r r pT

r r r p


Example: PUMA 560

After calculating the right side, it can been seen there’s a solvable solution:

And we get: -sin θ1 pX + cos θ1 pY = d3 .

a cosθi + b sin θi = c return 2 solutions:

1 1 11 12 13

1 1 21 22 23 16

31 32 33

0 0

0 0

0 0 1 0

0 0 0 1 0 0 0 1

X

Y

Z

c s r r r p

s c r r r pT

r r r p

2 2 2arctan 2( , ) arctan 2( , )i b a a b c c


Example: PUMA 560

2 solutions were found for θ1. For each one of them, we’ll continue to search for solvable equations…

1 1 11 12 13

1 1 21 22 23 16

31 32 33

0 0

0 0

0 0 1 0

0 0 0 1 0 0 0 1

X

Y

Z

c s r r r p

s c r r r pT

r r r p


Overview

KinematicsIntroduction to Protein Structure A kinematic View of Loop Closure


Overview

KinematicsIntroduction to Protein Structure

What do proteins look like and why is it important?

A kinematic View of Loop Closure


Proteins & PolypeptidesAmino Acids Polypeptides

Proteins


Where are they?

Proteins are very important molecules to all forms of life.

They are one of the four basic building blocks of life:– carbohydrates (sugars)– lipids (fats)– nucleic acids (DNA and RNA)– proteins


Where are they?

They serve all kinds of functions:– part of structural elements in a cell (small

scale)– part of the fibers that make up your muscles

(larger scale)– Enzymes– Antibodies– Hormones– ...


What are they?

Proteins are made up of a chain of amino acids linked together (peptide bonds).They may be seen as a chain (the backbone) with a lot of side chains (the residues).

There are 20 of those proteins' building blocks (amino acids)

What differs the amino acids is their residue.


Polypeptides and Proteins

Definition:A polypeptide is a compound containing amino acid residues joined by peptide bonds. A protein may consist of one or more specific polypeptide chains, which generally undergo further structural configurations in the course of becoming functional proteins.


Polypeptides and Proteins


Structure

Proteins have 4 increasingly complex levels of structure:– Primary: sequence of the amino acids– Secondary: common folding patterns seen in

proteins, like the alpha helix or the beta sheet– Tertiary: three-dimensional structure of a

single folded amino acid chain– Quaternary: the complete protein with all of

the subunits together (only in proteins made up of more than one polypeptide chain)


Structure


3D structure's importanceExample of the importance of the

structure:Boiling an egg causes all the proteins it contains, the "white" of the egg, to change shape and hardens (solid). It still has the same primary structure as the original protein, but the tertiary structure (three dimensional shape) has been lost, and so have all the critical properties of the original protein!


Structure's importanceProteins can have very complex shapes, and

the final form of the protein is essential to its intended function

The process of changing the shape of a protein so that the function is lost is called denaturation.


Structure-based problems

Docking: predicting whether (and how) one molecule will bind to another

Folding: the process by which the chain of amino acids is modified to reach its final form.

Loop Closure (see next)


Geometric Properties

Simplified (fixed bond lengths and angles)Planarity of the Peptide Bond:

A polypeptide chain (backbone) may be considered as a series of planes (peptide units) with two angles of rotation between each plane.



Each peptide bond adds 2 DOFs,



Each peptide bond adds 2 DOFs,and the residues too add some DOFs.


A Kinematic View of Loop Closure

Evangelos A. Coutsias,Chaok Seok,Matthew P. Jacobson,Ken A. Dill


Overview

KinematicsIntroduction to Protein StructureA kinematic View of Loop Closure


Overview

KinematicsIntroduction to Protein StructureA kinematic View of Loop Closure

redefining the loop closure problem as a “inverse kinematics” problem


Loop Closure

The loop closure problemTripeptide loop closureGeneralizations of the method


The loop closure problem

Definition:finding the ensemble of possible backbone structures of a chain segment of a protein molecule that is geometrically consistent with preceding and following part of the chain whose structures are given



missing

given a chain segmentof the protein

find possible backbonestructure

So that it is geometrically consistent with thepreceding and following parts of the chain



In his simplest form:given a molecular chain with inflexible bond length ad bond angles, find all possible arrangements with the property that all bond vectors are fixed in space except for a contiguous set and such that the changes are made in at most six intervening dihedral angles.


Protein bond with inflexible lengthinflexible bond angles


In his simplest form:

find all possible angles


Loop Closure



Tripeptide loop closure

The six-torsion loop closure problem in simplified representation:

fixed in space

variables: τi (i=1,2,3)constraints: θi (i=1,2,3)

τ1 τ2

τ3

θ1

θ2

θ3


Tripeptide loop closure

The six-torsion loop closure problem in simplified representation.But there are only 3 rotation angles τi ?

σi=τi+δi


Affixing frames


Affixing frames

τi angle is the rotation angle of riτ about

Zi


Affixing frames

σi angle is the rotation angle of ri σ about

Zi


Constraints

We’ve defined the frames so that they’ll be “easy to use”:– τi is defined by ri

τ

– σi is defined by ri σ

The θi angle constraints can be expressed in terms of ri

τ and ri σ:

riτ . ri

σ = cos θi


Constraints

riτ . ri

σ = cos θi

θi


Constraints

τi angle is the rotation angle of riτ about

Zi

Xi sin ηi

Zi cos ηi


Constraints

sin

0

cos

cos sin 0 sin

sin cos 0 0

0 0 1 cos

sin cos

sin sin

cos

i

i Z i

i

i i i

i i

i

i i

i i

i

r R


Constraints

Same for σi angle: is the rotation angle of ri

σ about Zi

1 1

1 1 1

1

sin cos

sin sin

cos

i i

i i i

i

r


Constraints


Constraints

Those equations describe the rotation of the Cαi-1-Ni and Cαi-Ci bonds about the virtual bonds Cαi-1-Cαi and Cαi-Cαi+1 respectively


Solving the equations

The angles αi, ηi, ξi, θi and δi depend on the bonds, which inflexible structure is known. Thus, they’re known constants.

Convert the equations to polynomial variables wi and ui:



The equations becomes:



Eliminating wi :

sin2

tan tan2 2

cos2

cos / 2 sin / 2 sin / 2 cos / 2

cos / 2 cos / 2 sin / 2 sin / 2

tan / 2 tan / 2

1 tan / 2 tan / 2 1

tan / 2

i i

i i ii

i i

i i i i

i i i i

i i i i

i i i i

i i

w

u

u

where



There are now 3 equations, quadratic in 2 variable (ui and ui+1)

Eliminating u1 and then u2 results in a degree 16 polynomial in u3

There might be up to 16 solutions, even if at most 10 real solutions has been found in the article’s research


Loop Closure



Generalizations of the method

There are no assumptions about the intervening structures. Therefore, the algorithm can be applied to moves involving arbitrary triads of Cα atoms.



Finding alternative local structures when an arbitrary dihedral angle is changed



The constraints form stay unchanged:

But, Cα1-Cα2, η1 and ξ2 changed. As a result, Zi (i=1,2) and αi (i=1,2,3) changed too.

Still, the equations can be derived with the changed parameters


References

Introduction to Robotics: Mechanics and Control (3rd Edition)John J. Craig

Kinematic View of Loop ClosureEvangelos A. Coutsias,Chaok Seok,Matthew P. Jacobson,Ken A. Dill

Cyclic Coordinate Descent: A Robotics Algorithm for Protein Loop ClosureAdrian A. Canutescu,Roland L. Dunbrack Jr.


Thank you

direct & inverse kinematics1 algorithmic robotics and motion planning (0368.4010.01) instructor:...

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direct inverse kinematics19

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