Intuitive Kinematics – Converting Between Forward and Reverse Definitions of Space

Intuitive Kinematics – Converting Between Forward and Reverse Definitions of Space

Lecture Series 2ME 4135 R. R. Lindeke

Intuitive Kinematics for Robot Manipulators Defining the concept of the Kinematic

Solution Finding Kinematic Solutions for

POSITIONAL issues (only!) Cartesian Manipulators Cylindrical Manipulators Spherical Manipulators Articulating Manipulators SCARA and Other Redundant

Manipulators

Intuitive Kinematics for Robot Manipulators Forward Kinematic Solutions:

Given the settings on each Joint (q1, q2, , qi) of the manipulator …

Determine the End Position (Xe,Ye, Ze)base achieved for the given structural size

Inverse Kinematic Solutions: Given Structural Size and an End (or Target)

Position (Xe,Ye, Ze)base … Determine the values for each Joint (q1, q2, , qi)

that can place the Manipulator there But, Note, this solution MAY NOT BE UNIQUE!

Cantilevered Cartesian Robot

P-P-P Configuration

J1

J3

J2

X0

Y0

Z0

Gantry Cartesian RobotP-P-P Configuration

X0Y0

Z0

J1

J2J3

FKS & IKS for a Cartesian Device In the Forward Sense: On the Gantry

J1 was at 435 cm J2 was at 283 cm J3 was at 199 cm and there is a COLLAPSE LENGTH of 75,

50, & 50 cm respectively Where is the End in the ‘Null (base)

Space’?

FKS -- Cartesian

Xe is J1 + Cl1 = 435 + 75 = 510 Ye is 283 + 50 = 333 Ze is 199 + 50 = 249 Exercise Care in the ordering of the

Joints as they relate to the Base Frame definition of Space!!!

Doing the IKS Given you want the End Position of:

(510, 333, 249) and the Collapse Lengths are as seen above (75, 50, & 50 respectively)

J1 is in the X0 direction therefore, J1 = 510 – 75 = 435 (as expected!) Similarly for J2 & J3 (which are 283 and 199

respectively) – again as expected Here because of the directions the Joint motions

follow and how the Base Axes are defined: J1 in the direction of X0; J2 in Y0; J3 in Z0 – but

be careful as we move deeper into robotics!

Cylindrical Robot Work EnvelopeP-R-P//R-P-P Configuration

Cylindrical Robot

Developing a FKS (model):

Given , Z & R Compute End

Position in terms of X0, Y0 and Z0 (Xe,Ye, Ze)

R

X0

Y0

Z0

FKS for Cylindrical Manipulator

( )*

( )*

( )

mach

mach

mach

mach R

mach R

mach Z

Xe R Cl Cos

Ye R Cl Sin

Ze Z Cl

IKS for Cylindrical Manipulator Here, before we go on let me state this:

Sine and Cosine inverses lead to ambiguous angles (they repeat each semi-circle) since they are built from a ratio of a signed over an unsigned vector (Y/R or X/R )

We MUST use inverse Tangent solutions to remove the ambiguity since tangent is built from a ratio of signed vectors (Y/X)

Cylindrical Robot

Doing IKS – given a value for:

X, Y and Z of End Compute , Z and

R

R

X0

Y0

Z0

IKS Cylindrical Manipulator2 2

3

*( / )*

2( , ):

ATan2 is a special form of InverseTangent that places the angle in the proper quadrant (I to IV) depending on the signs of Xe and Ye (the

R Xe Ye ClR SinYe Xe TanR Cos

ATan Xe Yewhere

2

denominatorand numerator of the ratio respectively)

machZ Ze Cl

Computing ATan2 Angles Atan2(Xvalue,Yvalue) is a special form of Tan-1 but computed to

retain quadrant identity Consider the 4-cases of: X = 8 and Y = 12

Atan2(8,12) = Tan-1(12/8) = 56.31 This is a 1st Quadrant angle!

Atan2(-8,12) = 90 + Tan-1(8/12) = 90 + 33.69 = 123.69 (alternatively it is 180 - Tan-1(12/8) = 180 – 56.31 = 123.69) This is a 2nd quadrant angle and not the -56.31 value that you

find when computing Atan with your calculator! Atan2(-8,-12) =180 + Tan-1(12/8) =180 + 56.31 = 236.31

This is a 3rd quadrant angle! Atan2(8,-12) = 270 + Tan-1(8/12) = 270 + 33.69 = 303.69

(alternatively it is 360 - Tan-1(12/8) = 360 – 56.31 = 303.69) This is a 4th quadrant angle which is the same as your

calculator gives you (-56.31)!

Spherical Robot WorkspaceR-R-P Configuration

Spherical Robot

Developing a FKS (model):

Given , & R Compute Xe, Ye, &

Ze

R

X0

Y0

Z0

FKS Spherical Manipulator

0 0R' = projection of Radial Arm to X --Y plane' *

'*'*

*

given

given

R R Cos

Xe R CosYe R SinZe R Sin

Spherical Robot

Developing a IKS (model):

Given Xe, Ye, & Ze Compute , & R

R

X0

Y0

Z0

IKS, Spherical Manipulator

2 2 23

2 2

2 2

R = XeAgain, for Angles we desire :

''

tan 2( , )

'

tan 2( ', ):

':

tan 2 ,

: can be thought of

Ye Ze Cl

Ye R Sin TanXe R Cos

A Xe YeZe RSin TanR RCos

A R Zewhere

R Xe Yeso

A Xe Ye Ze

note

Tangent Inverses

2 as

2-Link Articulating Arm Manipulator

3

1

2

L2

L1

• This Machine Rotates about Z0 Axis (1)

• 2 is measured RELATVE to the base plane

• 3 is measured RELATVE to L1 not the base plane

• All Joint Angles are Right Hand Rule Based

Z0

X0Y0

FKS 2-Link A. ARM Manipulator

1 1 2 2 2 3

1 1 2 2 2 3

1 2 2 2 3

* * *

* * *

* *

Xe Cos L Cos L Cos

Ye Sin L Cos L Cos

Ze L Sin L Sin

2-Link Articulating Arm Manipulator

3

12

L2

L1

Z0

X0Y0

IKS 2-Link A. Arm All angles defined as ATan2 (using

target End Position & Link Lengths) Requires Construction Lines!! First Solve for 1 = ATan2(Xe, Ye) Then the Tilt Angles (3 & 2) in

reversed Order (as stated)! Solution Indicates 2 acceptable

configurations for the Arm: ‘Elbow UP’ and ‘Elbow DOWN’

IKS 2-Link A. Arm1

1

0 0

3 2

tan 2( , )Rotate Arm to Position andDefine a Plane normal to X -Y andContaining the Arm LinksDefine a Vector R from origin of newPlane to End of ArmSolve for and

A Xe Ye

IKS Q2 & Q3

2

3

a

IKS 2-Link Art. Arm (3)

2 2 21 2 1 2 3

2 2 2 2

3 3

3

2 2 21 2

321

33

3

2 23 3

2

3

23

Using Law of Cosines:

2 (180 ):

:(180 )

solving for COS :

2

:

1 1

1

2 , 1

R L L L L Coswhere

R Xe Ye Ze

andCos Cos

R L LCos DL L

SinTan Cosand

Sin Cos D

DTan D

ATan D D

IKS 2-Link Art. Arm (2)

00

1

2

2 2

2 2 22 1 1

2 2 21 2

1

Lets define the angle from the X -Y planeto Vector R and as the angle between L and R

:

tan 2 ,

Solving for :Law of Cosines:

2 ( )

( ) 2Law o

Therefore

A Xe Ye Ze

L R L RLCos

R L LCos RL

a

a

a

a

a

3

2

3 3

2 3

f Sines:( ) (180 )

:(180 )

( )

Sin SinL R

hereSin Sinso

L SinSin R

a

a

IKS 2-Link A. Arm (2) cont.

2 3

2 2 21 2

1

1 2 32 2 2

1 2

2 2 21 2 1 2 3

( )( ) ( )

( )

2:2( )

2 , 2

SinTan CosL Sin

RTanR L L

RLsimplifying

L L SinTanR L L

ATan R L L L L Sin

aa a

a

a

a

IKS 2-Link A. Arm (2) cont.

2 2 2 2 22 1 2 1 2 3

23

2 2 2 2 2 22 1 2 1 2

2 2 21 2

21

:

tan 2 , 2 , 2

Re

1

tan 2 , 2 , 2 * 1

:

2

Finally

A Xe Ye Ze ATan R L L L L Sin

membering

Sin D

A Xe Ye Ze ATan R L L L L D

where

R L LDL L

a

SCARA Manipulator – an over specified Planer Articulating Arm

FKS & IKS for this over specified Arm – Only 2 “State Equations” Exist for 3 Variables

In a Forward Sense solution is reasonable

In an Inverse sense the above statement indicates the existence of an infinite number of possible solutions

FKS: Given 1,2,3 find Xe and Ye

IKS: Given Xe and Ye (& L1, L2 and L3) find: 1,2,3

FKS 3-Link Planer A. Arm

y1

y2

y3

x1 x2 x3

1

2

3

Focusing on the FKS Project each link to the two axes (X & Y) Xe is found by summing X-projected

lengths of each link Ye found by summing Y-projected lengths

of the links Example:

Link 1 to X: L1*Cos(1) Link 1 to Y: L1*Sin(1)

FKS Continuing X2 is projection of L2 – to the X axis therefore it

requires what Projection Factor? Sure, Cos(1+ 2) Y2 is projection of L2 to the Y axis so it is equal

to: L2* Sin(1+ 2) Finally:

Xe = L1*Cos(1) + L2*Cos(1+ 2) + L3*Cos(1+ 2+ 3)

Ye = L1*Sin(1) + L2*Sin(1+ 2) + L3*Sin(1+ 2+ 3)

What about the IKS It’s a 2 Step Process Requires a Parameterization of one of

the Joint Angles – This step will establish the acceptable limits of the solution space

The Parameterized Joint is said to set the BOUNDS for the solution space

We select Joint 1 as the one to be parameterized

3-Link Planer Arm IKS – Step 1

1

2’

Note: L2’ is the sum of L2 + L3 formed by freezing 3 at 0˚

1 Is Solved as Above It is the lower angle of a 2-link

Articulating Arm The 2 solutions found for 1 thus form

the Upper and Lower Bounds for the solution space

Now pick one within the range Using this angular value, “Transform”

the solution space up Link L1 and ‘unfreeze’ 3

IKS Step 2:’Redefine’ Space at end of L1 and Re-apply 2-link method in the Transformed space

1,picked

Homework Assignment: Complete the IKS solution for a 3-link

planar articulating arm Consider and complete the FKS and

IKS solution for a “Planer” P-R-P device