19AIE201

Introduction to Robotics

IRB 140

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

IRB 140

Text

" Small, powerful, and fast 6-axes robot. Compact and very powerful. The IRB 140 six axes multipurpose industrial robot handles a payload of 6 kg with a reach of 810 mm (to axis 5) It can be floor mounted, inverted or wall mounted at any angle. The robust design with fully integrated cables adds to the overall flexibility and the collision detection function ensures the robot is reliable and safe."

\text{Rotation about } X_{i-1} \rightarrow \alpha_{i-1}

Link Twist

Z

Link Length

Z

\text{Translation along } X_{i-1} \rightarrow a_{i-1}

Joint Angle

X

\text{Rotation about } Z_{i} \rightarrow \theta_{i}

Link offset

X

\text{Translation along } Z_{i} \rightarrow d_{i}

\text{Rotation about } X_{i-1} \rightarrow \alpha_{i-1}

\text{Rotation about } Z_{i} \rightarrow \theta_{i}

\text{Translation along } Z_{i} \rightarrow d_{i}

\text{Translation along } X_{i-1} \rightarrow a_{i-1}

MDH Parameters


0	0	0.352
-90	0.070	0
0	0.360	0
-90	0	0.380
90	0	0
-90	0	0

\alpha_{i-1}

a_{i-1}

d_{i}

\theta_{i}

\theta_{1}

\theta_{2}-90

\theta_{3}

\theta_{4}

\theta_{5}

\theta_{6}

FORWARD AND INVERSE KINEMATICS ANALYSIS AND VALIDATION OF THE ABB IRB 140 INDUSTRIAL ROBOT

Forward Kinematics

\text{Joint angles} \rightarrow \text{End effector position}

T^0_6 = Tmdh(0,1)\cdot Tmdh(1,2)\cdot Tmdh(2,3)\cdot Tmdh(3,4)\cdot Tmdh(4,5)\cdot Tmdh(5,6)

MDH Parameters - Given

T^0_6 \rightarrow F(\theta_1,...,\theta_6)

MDH Parameters - Given

T^0_6 \rightarrow F(\theta_1,...,\theta_6)

{\theta_1,...,\theta_6} \rightarrow Input

T^0_6 \rightarrow F(\theta_1,...,\theta_6)

{\theta_1,...,\theta_6} \rightarrow Input

Inverse Kinematics

\text{End effector position} \rightarrow \text{Joint angles}

T^0_6 \rightarrow Given

MDH Parameters - Given

T^0_6 \rightarrow Given

MDH Parameters - Given

t^0_6 \rightarrow obtained

t^0_6 =T^0_6

How to solve this??

Before that, lets see an application based on IK

Pantograph

19AIE201

Introduction to Robotics

Closed form solution

Closed Form Solution

\theta_1 = atan2(y,x)

\theta_1 = \pi + atan2(y,x)

Closed Form Solution

cos(\zeta) = \frac{s^2 + r^2 - a_2^2 - (d_4 + d_6)^2}{2 a_2 (d_4 + d_6)}

sin(\zeta) = \pm \sqrt{1 - cos^2(\zeta)}

\zeta = atan2(sin(\zeta) , cos(\zeta))

\theta_3 = -(\frac{\pi}{2} + \zeta)

Closed Form Solution

\theta_3 = -(\frac{\pi}{2} + \zeta)

\theta_2 = atan2(s,r) - atan2((d_4 + d_6)sin(\zeta) , a_2 + (d_4 + d_6)cos(\zeta))

\theta_3 = -(\frac{\pi}{2} + \zeta)

\theta_2 = atan2(s,r) - atan2((d_4 + d_6)sin(\zeta) , a_2 + (d_4 + d_6)cos(\zeta))

\theta_1 = atan2(y,x)

\theta_4, \theta_5 , \theta_6 \text{ can be found by comparing to}\\ R^3_6

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

19AIE201

Introduction to Robotics

Trajectory Planning

Tracking the change in position, velocity and acceleration of joints w.r.t the time.

Trajectory Planning

Polynomial Function

Trajectory Planning

Polynomial Function

\theta(t) = a_0 + a_1t + a_2t^2 + a_3t^3

Trajectory Planning

Polynomial Function

\theta(t) = a_0 + a_1t + a_2t^2 + a_3t^3

\dot{\theta(t)} = a_1 + 2a_2t + 3a_3t^2

Trajectory Planning

Polynomial Function

\theta(t) = a_0 + a_1t + a_2t^2 + a_3t^3

\dot{\theta(t)} = a_1 + 2a_2t + 3a_3t^2

\ddot{\theta(t)} = 2a_2 + 6a_3t

Trajectory Planning

Constraints

Trajectory Planning

Constraints

\theta(0) = \theta_0

Trajectory Planning

Constraints

\theta(t_f) = \theta_f

\theta(0) = \theta_0

Trajectory Planning

Constraints

\theta(t_f) = \theta_f

\theta(0) = \theta_0

\dot{\theta(0)} = 0

Trajectory Planning

Constraints

\theta(t_f) = \theta_f

\theta(0) = \theta_0

\dot{\theta(0)} = 0

\dot{\theta(t_f)} = 0

Trajectory Planning

Using these constraints...

\theta(t_f) = \theta_f

\theta(0) = \theta_0

\dot{\theta(0)} = 0

\dot{\theta(t_f)} = 0

\theta(t) = a_0 + a_1t + a_2t^2 + a_3t^3

\dot{\theta(t)} = a_1 + 2a_2t + 3a_3t^2

\ddot{\theta(t)} = 2a_2 + 6a_3t

Trajectory Planning

Using these constraints...

\theta(t_f) = \theta_f

\theta(0) = \theta_0

\dot{\theta(0)} = 0

\dot{\theta(t_f)} = 0

\theta(t) = a_0 + a_1t + a_2t^2 + a_3t^3

\dot{\theta(t)} = a_1 + 2a_2t + 3a_3t^2

\ddot{\theta(t)} = 2a_2 + 6a_3t

\Rightarrow

Trajectory Planning

In these equations...

\theta(t_f) = \theta_f

\theta(0) = \theta_0

\dot{\theta(0)} = 0

\dot{\theta(t_f)} = 0

\theta(t) = a_0 + a_1t + a_2t^2 + a_3t^3

\dot{\theta(t)} = a_1 + 2a_2t + 3a_3t^2

\ddot{\theta(t)} = 2a_2 + 6a_3t

\Rightarrow

Trajectory Planning

We get this

a_0 = \theta_0

a_1 = 0

a_2 = \frac{3}{t^2}(\theta_f - \theta_0)

a_3 = -\frac{2}{t^3}(\theta_f - \theta_0)

19AIE201

Introduction to Robotics

Jacobian

Jacobian

x = f_1(\theta_1 ,\theta_2 ,\theta_3 ,\theta_4 ,\theta_5 ,\theta_6 )\\ y = f_2(\theta_1 ,\theta_2 ,\theta_3 ,\theta_4 ,\theta_5 ,\theta_6 )\\ z = f_3(\theta_1 ,\theta_2 ,\theta_3 ,\theta_4 ,\theta_5 ,\theta_6 )

J_v

Jacobian

\omega_i = \dot{q_i}\hat{z_i}

J_{\omega}

Jacobian

J = \begin{bmatrix} J_v & J_{\omega} \end{bmatrix}

\begin{bmatrix} v_x \\ v_y\\ v_z\\ \omega_x\\ \omega_y\\ \omega_z \end{bmatrix} = J\cdot \vec{\dot{q}}