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3. You are presented with the RRR manipulator below. L1, L2, and L3 are strictly positive. Z3 ?1 ?2 ?3 Z4 X4 L? L2 L3 Z1, Zo (out) Z2 (out) (a) (3 Points) Find the Denavit-Hartenberg parameters for this manipulator. Assign the frames such that all your a? are positive. i a??? ???? d? ?? 1 2 3 4 (b) (3 Points) The position of the end-effector is: $\begin{pmatrix} L_1c_1 + L_2c_{12} + L_3c_{12}c_3 \\ L_1s_1 + L_2s_{12} + L_3s_{12}c_3 \\ -L_3s_3 \end{pmatrix}$, where $c_{12} = \cos(\theta_1 + \theta_2)$. Derive the linear Jacobian $^0J_v$. (c) (6 Points) Find the singular configurations of this manipulator. For each singularity, draw the robot configuration and clearly state how the movement is restricted (in terms of frame axes). Hint: The linear Jacobian in frame {2} is given to you here: $\begin{pmatrix} -L_1s_2 & 0 & -L_3s_3 \\ L_1c_2 + L_2 + L_3c_3 & L_2 + L_3c_3 & 0 \\ 0 & 0 & -L_3c_3 \end{pmatrix}$

          3. You are presented with the RRR manipulator below. L1, L2, and L3 are strictly positive.
Z3
?1
?2
?3
Z4
X4
L?
L2
L3
Z1, Zo
(out)
Z2
(out)
(a) (3 Points) Find the Denavit-Hartenberg parameters for this manipulator. Assign the frames such
that all your a? are positive.
i a??? ???? d? ??
1
2
3
4
(b) (3 Points) The position of the end-effector is:
$\begin{pmatrix} L_1c_1 + L_2c_{12} + L_3c_{12}c_3 \\ L_1s_1 + L_2s_{12} + L_3s_{12}c_3 \\ -L_3s_3 \end{pmatrix}$,
where $c_{12} = \cos(\theta_1 + \theta_2)$.
Derive the linear Jacobian $^0J_v$.
(c) (6 Points) Find the singular configurations of this manipulator. For each singularity, draw the
robot configuration and clearly state how the movement is restricted (in terms of frame axes).
Hint: The linear Jacobian in frame {2} is given to you here:
$\begin{pmatrix} -L_1s_2 & 0 & -L_3s_3 \\ L_1c_2 + L_2 + L_3c_3 & L_2 + L_3c_3 & 0 \\ 0 & 0 & -L_3c_3 \end{pmatrix}$

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3. You are presented with the RRR manipulator below. L1, L2, and L3 are strictly positive.
Z3
?1
?2
?3
Z4
X4
L?
L2
L3
Z1, Zo
(out)
Z2
(out)
(a) (3 Points) Find the Denavit-Hartenberg parameters for this manipulator. Assign the frames such
that all your a? are positive.
i a??? ???? d? ??
1
2
3
4
(b) (3 Points) The position of the end-effector is:
< p m a t r i x >,
where c12 = cos(θ1 + θ2).
Derive the linear Jacobian ^0Jv.
(c) (6 Points) Find the singular configurations of this manipulator. For each singularity, draw the
robot configuration and clearly state how the movement is restricted (in terms of frame axes).
Hint: The linear Jacobian in frame 2 is given to you here:
< p m a t r i x >

Added by Justin G.

Computer Science and Information Technology

Trishna Knowledge Systems 2018 Edition

Instant Answer

Solved by Expert Adi S

07/30/2023

Step 1

The parameters are: * **a_i-1**: The distance between the z_i-1 and z_i axes along the x_i-1 axis. * **alpha_i-1**: The angle between the z_i-1 and z_i axes about the x_i-1 axis. * **d_i**: The distance between the x_i-1 and x_i axes along the z_i axis. * Show more…

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3. You are presented with the RRR manipulator below. L1, L2, and L3 are strictly positive. Z3 01 02 03 Z4 X4 L₁ L2 L3 Z1, Zo (out) Z2 (out) (a) (3 Points) Find the Denavit-Hartenberg parameters for this manipulator. Assign the frames such that all your a₁ are positive. i Ai-1 0-1 di Oi 1 2 3 4 (b) (3 Points) The position of the end-effector is: L1C1+L2C12+ L3C12C3 P4 = L181 + L2812+ L3812C3 -L353 where C12 = cos(01+02). Derive the linear Jacobian Ju. (c) (6 Points) Find the singular configurations of this manipulator. For each singularity, draw the robot configuration and clearly state how the movement is restricted (in terms of frame axes). Hint: The linear Jacobian in frame {2} is given to you here: -L182 0-L383 2J = L1C2+L2+L3C3 L2+L3C3 0 0-L3C3

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Transcript

00:02 The dh parameter for the given manipulator is linked 1, 2, 3.

00:17 A1's are 0, 0, 0.

00:22 Alpha i, 0, minus 90, 0.

00:30 Di, d1, d2, d3.

00:37 And theta i's are theta 1, 0, 0.

00:42 Therefore, a1 is equal to matrix cos theta 1 minus sin theta 1, 0, 0.

00:59 Sin theta 1, cos theta 1, 0, 0.

01:08 0, 0, 1, d1.

01:13 0, 0, 0.

01:20 And a2 is equal to matrix of 1, 0, 0, 0.

01:31 0, 0, 1, 0.

01:35 0, minus 1, 0, d2.

01:42 0, 0, 0.

01:44 And a3 is given as matrix 1, 0, 0, 0.

01:57 0, 1, 0, 0.

02:00 0, 0, 1, d3.

02:05 0, 0, 0, 1.

02:09 Therefore, the jacobi matrix a3, 0 is equal to a1 multiplied by a2, multiplied by a3.

02:18 Therefore, a1 multiplied by a2 is equal to matrix cos theta 1 minus sin theta 1, 0, 0.

02:32 Sin theta 1, cos theta 1, 0, 0.

02:40 And 0, 0, 1, d1.

02:45 And 0, 0, 0, 1.

02:51 And a3 is given as matrix of 1, 0, 0, 0...

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