Homework 4

Course	Introduction to Robotics
Course Number	00350001

סטודנט א’	סטודנט ב’	סטודנט ג’
עידו פנג בנטוב	ניר קרל	אופיר רובין
CLASSIFIED	CLASSIFIED	CLASSIFIED
CLASSIFIED	CLASSIFIED	CLASSIFIED

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Question 1

We are given:

Figure HW4.1: Given manipulator.

Therefore the world to frame transformations:

Part a

To compute the Jacobian matrix, we’ll first follow the systematic algorithm, and as an alternative we’ll look at the shortcut.

Systematic Algorithm:
From the transformation matrices, we extract:

For frame :

For frame :

For frame :

For frame :

Now to compute the Jacobian columns, for joint (revolute joint):

For joint (revolute joint):

For joint (revolute joint):

Therefore:

Shortcut:
From the forward kinematics, the end-effector position is:

We can compute the linear Jacobian by direct differentiation:

Part b

The joint values at which the manipulator configuration is singular are where . That is:

Therefore, the manipulator is singular when:

These correspond to:

1. : The second and third joints are collinear
2. : The end-effector is at maximum/minimum reach in the -direction

Figure HW4.2: (Left) Singular configuration where . (Right) Singular configuration where .

Part c

The new manipulator will be of the form:

Figure HW4.3: RR manipulator with fixed.

For the RR manipulator with fixed, we have and , so and .

The end-effector position becomes:

The Jacobian matrix for this RR manipulator is:

This is a matrix. For the manipulator to be non-singular, this matrix must have full rank (rank = ).

Singularity Analysis:
The rank is less than when all minors are zero. The possible minors are:

Minor from rows (1,2):

Minor from rows (1,3):

Minor from rows (2,3):

For singularity, ALL minors must be zero simultaneously.

From and :
Both minors contain the factor . Since , we need (i.e., ).

Since when , not all minors are zero, so is NOT a singularity.

From :

This requires either:

1. (i.e., ), or

Case 1: ()

• When : , so:

  • (unless )
  • (unless )

  Since and cannot both be zero, at least one of or is non-zero.

• When : , so:

For ALL minors to be zero at , we need both and . Since and cannot both be zero, we need , i.e., .

Case 2:
This gives . For this to be valid, we need .

Singularity Condition:
The RR manipulator is singular when:

This includes the special case when .

Question 2

We are given:

Figure HW4.4: Given manipulator.

Part a

The linear Jacobian would be:

Therefore the whole Jacobian:

Where since this is a planar manipulator.

Part b

The joint values at which the manipulator configuration is singular are where the linear Jacobian loses rank. Since this is a planar manipulator with 3 DOF but only 2 spatial dimensions, we need to check when all minors of become zero.

The three possible minors are:

Minor from columns (1,2):

Minor from columns (1,3):

Minor from columns (2,3):

Singularity Conditions:
The manipulator is singular when any of these minors equals zero:

1. From :
2. From : , which gives
3. From : , which gives

Therefore, the complete set of singularities is:

Figure HW4.5: (Left) Singular configuration where . (Center) Singular configuration where . (Right) Singular configuration where .

Part c

Given that masses and are located at the middle of their respective links, and mass is held at the gripper’s origin, we need to compute the generalized force vector required to maintain static equilibrium under gravity.

Solution:

Using the principle of superposition from manipulator statics, we treat each mass separately and sum their contributions:****

Determine Mass Positions

From the transformation matrix and manipulator configuration:

Mass (middle of link 1):
Link 1 extends from the base to joint 2 over distance . The center of mass is at:

Mass (middle of link 2):
Link 2 is the prismatic joint extending distance from joint 1. The center of mass is at:

Mass (at gripper origin):
From the given transformation matrix:

Compute Jacobians for Each Mass

For mass :
Only joint affects the position of :

For mass :
Joints and affect the position of . Note that for the prismatic joint (joint 2), we use the force transmission vector rather than the kinematic derivative, since the joint must support the full gravitational force of the distributed mass:

For mass :
We use the full Jacobian from part a (linear part only for planar motion):

Apply Gravitational Forces

Each mass experiences gravitational force:

Compute Joint Torques

Using :

For mass :

For mass :

For mass :

Total Generalized Forces

Summing all contributions:

For each joint:

The generalized forces that gravity applies to the system are:

Therefore, the generalized force vector that the joints must apply to maintain static equilibrium is :