19AIE201

Introduction to Robotics

Evaluation 6

19AIE201

Introduction to Robotics

Aadharsh Aadhithya - CB.EN.U4AIE20001
Anirudh Edpuganti - CB.EN.U4AIE20005

Madhav Kishor - CB.EN.U4AIE20033

Onteddu Chaitanya Reddy - CB.EN.U4AIE20045
Pillalamarri Akshaya - CB.EN.U4AIE20049

Team-1

19AIE201

Introduction to Robotics

Pieper's Solution for 6-axis robots

19AIE201

Introduction to Robotics

Piper's Solution for 6-axis robots

Conditions

Last three axes of manipulator must be intersecting
All the six joints must be revolute joints

19AIE201

Introduction to Robotics

Piper's Solution for 6-axis robots

19AIE201

Introduction to Robotics

Piper's Solution for 6-axis robots

The distance between the axes 1,2,3 and axes 4,5,6 only depends on
The height of end manipulator only depends on and

\theta_3

\theta_2

\theta_3

19AIE201

Introduction to Robotics

Piper's Solution for 6-axis robots

\begin{bmatrix} a_3 \\ -sin(\alpha_3)d_4\\ cos(\alpha_3)d_4\\ 1 \\ \end{bmatrix} \quad

{^3P_{4ORG}} =

where \ {^AP_{Borg}} \ is \ the \ origin\ of \ frame \ B \ expressed \ in \ coordinate \ of \ frame \ A

\begin{bmatrix} a_3 \\ -sin(\alpha_3)d_4\\ cos(\alpha_3)d_4\\ 1 \\ \end{bmatrix} \quad

{^0P_{4ORG}} =

^2_3T

^1_2T

^0_1T

19AIE201

Introduction to Robotics

Piper's Solution for 6-axis robots

^2_3T = \begin{bmatrix} cos(\theta_3) & -sin(\theta_3) & 0 & a_2 \\ sin(\theta_3)cos(\alpha_2) & cos(\theta_3)cos(\alpha_2) & -sin(\alpha_2) & -d_3sin(\alpha_2) \\ sin(\theta_3)sin(\alpha_2) & cos(\theta_3)sin(\alpha_2) & cos(\alpha_2) & d_3cos(\alpha_2) \\ 0 & 0 & 0 & 1 \end{bmatrix}

\vec{F} = \begin{bmatrix} f_1(\theta_3) \\ f_2(\theta_3) \\ f_3(\theta_3) \\ 1 \end{bmatrix} = \ ^2_3T \begin{bmatrix} a_3 \\ -sin(\alpha_3)d_4\\ cos(\alpha_3)d_4\\ 1 \\ \end{bmatrix} \quad

19AIE201

Introduction to Robotics

Piper's Solution for 6-axis robots

^2_3T = \begin{bmatrix} cos(\theta_3) & -sin(\theta_3) & 0 & a_2 \\ sin(\theta_3)cos(\alpha_2) & cos(\theta_3)cos(\alpha_2) & -sin(\alpha_2) & -d_3sin(\alpha_2) \\ sin(\theta_3)sin(\alpha_2) & cos(\theta_3)sin(\alpha_2) & cos(\alpha_2) & d_3cos(\alpha_2) \\ 0 & 0 & 0 & 1 \end{bmatrix}

\vec{F} = \begin{bmatrix} f_1(\theta_3) \\ f_2(\theta_3) \\ f_3(\theta_3) \\ 1 \end{bmatrix} = \ ^2_3T \begin{bmatrix} a_3 \\ -sin(\alpha_3)d_4\\ cos(\alpha_3)d_4\\ 1 \\ \end{bmatrix} \quad

19AIE201

Introduction to Robotics

Piper's Solution for 6-axis robots

\vec{G} = \begin{bmatrix} g_1(\theta_2, \theta_3) \\ g_2(\theta_2, \theta_3) \\ g_3(\theta_2, \theta_3) \\ 1 \end{bmatrix} = \ ^1_2T \begin{bmatrix} f_1(\theta_3) \\ f_2(\theta_3) \\ f_3(\theta_3) \\ 1 \end{bmatrix}

^1_2T = \begin{bmatrix} cos(\theta_2) & -sin(\theta_2) & 0 & a_1 \\ sin(\theta_2)cos(\alpha_1) & cos(\theta_2)cos(\alpha_1) & -sin(\alpha_1) & -d_2sin(\alpha_1) \\ sin(\theta_2)sin(\alpha_1) & cos(\theta_2)sin(\alpha_1) & cos(\alpha_1) & d_2cos(\alpha_1) \\ 0 & 0 & 0 & 1 \end{bmatrix}

19AIE201

Introduction to Robotics

Piper's Solution for 6-axis robots

^2_1T = \begin{bmatrix} cos(\theta_2) & -sin(\theta_2) & 0 & a_1 \\ sin(\theta_2)cos(\alpha_1) & cos(\theta_2)cos(\alpha_1) & -sin(\alpha_1) & -d_2sin(\alpha_1) \\ sin(\theta_2)sin(\alpha_1) & cos(\theta_2)sin(\alpha_1) & cos(\alpha_1) & d_2cos(\alpha_1) \\ 0 & 0 & 0 & 1 \end{bmatrix}

^0_1T = \begin{bmatrix} cos(\theta_1) & -sin(\theta_1) & 0 & 0 \\ sin(\theta_1) & cos(\theta_1) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}

{^0P_{4ORG}} = \begin{bmatrix} cos(\theta_1)g_1(\theta_2,\theta_3) - sin(\theta_1)g_2(\theta_2,\theta_3) \\ sin(\theta_1)g_1(\theta_2,\theta_3) + cos(\theta_1)g_2(\theta_2,\theta_3) \\ g_3(\theta_2,\theta_3)\\ 1 \\ \end{bmatrix} \quad

19AIE201

Introduction to Robotics

Piper's Solution for 6-axis robots

Case \ Number \ 1 \: \ a_1 = 0 \\ This\ will \ be \ the\ case \ when \ the \ axes \ 1 \ and \ 2 \ intersect

r^2=t^2_{14}+t^2_{24}+t^2_{34}=f^2_1(\theta_3)+f^2_2(\theta_3)+f^2_3(\theta_3)+d^2_2+2d_2f_3(\theta_3)

Simplifying \ the \ above \ equation \ we \ get \ the \ value \ of \ \theta_3 \\ Subsutituting \ the \ value \ of \ \theta_3 \ in \ the \ equations \ we \ get \ \theta_1 \ \theta_2

^3_6T_{des} = ^0_3T^{-1} \ ^0_6T_{des}

Now \ solving \ the \ above \ matrix \ we \ get \ the \ values \ of \ \theta_4, \ \theta_5, \ \theta_6

19AIE201

Introduction to Robotics

Piper's Solution for 6-axis robots

Case \ Number \ 2 \: \ \alpha_1 = 0 \\ The \ height \ of \ the \ spherical \ wrist \ centre \ in \ the \ 0 \ frame \ will \ be \

g_3(\theta_2,\theta_3) = sin(\theta_2)sin(\alpha_1)f_1(\theta_3) + cos(\theta_2)sin(\alpha_1)f_2(\theta_3) + cos(\alpha_1)f_3(\theta_3) + d_2cos(\alpha_1)

g_3(\theta_3) = f_3(\theta_3) + d_2

Simplifying \ the \ above \ equation \ we \ get \ the \ value \ of \ \theta_3 \\ Subsutituting \ the \ value \ of \ \theta_3 \ in \ the \ above \ equation \ we \ get \ \theta_1 \ \theta_2

^3_6T_{des} = ^0_3T^{-1} \ ^0_6T_{des}

Now \ solving \ the \ above \ matrix \ we \ get \ the \ values \ of \ \theta_4, \ \theta_5, \ \theta_6

19AIE201

Introduction to Robotics

Newton-Raphson method

19AIE201

Introduction to Robotics

Newton-Raphson method

^0_6T_{\theta} = \ ^0_6T(\theta_1,\theta_2,\theta_3,\theta_4,\theta_5,\theta_6)

^0_6T_{des} \ is \ a \ final \ orientation \ matrix \ of \ end \ effector \\

Let \ ^0_6T_{\theta}(a,b) \ denote \ the \ element \ of \ matrix \ ^0_6T_{\theta} \ that \ is \ in \ the \ ath \ row \ and \ bth \ column

\theta = \ ^0_6T_{\theta} - ^0_6T_{des}

Newton-Raphson method

\theta_{n+1} = \theta_n - \frac{F(\theta_n)}{F'(\theta_n)}

Newton-Raphson method

\theta_{n+1} = \theta_n - \frac{F(\theta_n)}{F'(\theta_n)}

\theta_{n+1} = \theta_n - J(\theta_n)^{-1}F(\theta_n)

Newton-Raphson method

\theta_{n+1} = \theta_n - J(\theta_n)^{-1}F(\theta_n)

Jacobian Matrix