Chapter1. Intro and Preliminaries

Given by: Dr. Hamid Sadeghian

Geometric Jacobian

To express

• end-effector linear velocity $\dot{p}_e = J_P(q)\dot{q}$
• angular velocity $\omega_e = J_O(q)\dot{q}$
• velocity

$$\begin{bmatrix} \dot{p}_e \\ \omega_e \end{bmatrix} = J(q)\dot{q}$$

• Jacobian

$$J = \begin{bmatrix} J_P \\ J_O \end{bmatrix}$$

as a function of joint velocities $\dot{q}$

The column of Jacobian $J_i$ represents:

The effect on the end-effector velocity when the i th joint moves independently at a unit velocity if a robot has n dof

$$J \in \mathbb{R}^{6 \times n}$$

Redundancy and optimization

Jacobian acts as a mapping from the space of joint velocities to the space of task velocities. The optimal solution is given by

$$\dot{q} = W^{-1}J^T(JW^{-1}J^T)^{-1}v_c$$

where

$$J_W^\dagger = W^{-1}J^T (JW^{-1}J^T)^{-1}$$

Jacobian 的 weighted pseudoinverse(加权伪逆). W is the weight matrix, usually symmetric and positive definite.

example:

$$W= \begin{bmatrix} 1 & 0 \\ 0 & 10 \end{bmatrix}$$

The second joint movement is more expensive.

Why we need weighted pseudoinverse

for redundant robot $n>6$, joints are more than task domain, we have infinate $\dot{q}$ to represent a $v_e$, so we need to have the optimal from infinate solutins.

$$\min_{\dot{q}} \quad \frac12 \dot{q}^T W \dot{q}$$

If W = I, then it is the normal Moore-Penrose pseudoinverse:

$$J^\dagger = J^T(JJ^T)^{-1}$$

Weighted pseudoinverse can be understand as

Which should move less

• joint close to limit
• joint energy cost is high
• joint has poor precision
• joint is not needed Then the weight is higher, the system will use low cost joints for higher priority and dont use expensive joints.

$$\min_{\dot{q}} \frac12 \dot{q}^T W \dot{q}$$

s.t. optimal solution minimize the norm of joint velocities

$$J\dot{q}=v_e$$ $$\dot{q} = v_eJ^{\dagger}$$

it is from constrained optimization, it is weighted least norm solution。

$$\dot{q} = W^{-1}J^T (JW^{-1}J^T)^{-1} v_e$$

Null space control

if $\dot{q}^*$ is a solution, $\dot{q}^*+ P\dot{q_0}$ is also a solution, thus the general solution will be

$$\dot{q} = J^{\dagger}v_e + P\dot{q}_0$$

where P is the projection matrix to given by $P = (I-J^{\dagger}J)$ The inverse kinematic solution to follow a given task space trajectory is generalized to $\dot{q} = J^{\dagger}(\dot{x}_d + Ke)+(I_n-J^{\dagger}J)\dot{q}_0$, where $e = x_d - x$. The null-space velocity vector $\dot{q}_0$ can be used to minimize some objective function w(q) by choosing

$$\dot{q}_0 = k_0 \left( \frac{\partial w(q)}{\partial q} \right)^T$$

where $w(q)$ is objective function and gradient pointing to the fastest growing direction.

Main task is $\dot{q} = J^{\dagger}(\dot{x}_d + Ke)$ which helps the end-effector follows the desired trajectory. e is the desired pose. Ke is the feedback corrrection, pulls the robot back to the desired trajectory.

Null-space motion $(I_n-J^{\dagger}J)\dot{q}_0$ is satisfied and is not influenving the end-effector. Which keeps the joint motions that not influcing the main task. Why? Because

$$J(I_n-J^\dagger J)=0$$

objective functions

• The manipulability measure, $w(q) = \sqrt{\det(J(q)J^T(q)}$
• The distance from mechanical joint limits, $q_i$ is current joint angle,$\bar{q}_i$ is joint range

$$w(q) = - \frac{1}{2n} \sum_{i=1}^{n} \left( \frac{q_i-\bar q_i} {q_{iM}-q_{im}} \right)^2$$

let joint stay in rhe middle, not close to joint limits. System is maximize w(q)

• The distance from an obstacle, defined as,

$$w(q) = \min_{p,o} |p(q)-o|$$

system will increase the distance from obstacle.