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Velocity, Force and the Jacobian

Velocity for A Single Joint

Prismatic Joint

v=q˙rω=0\begin{align} v &= \dot{q} r \\ \omega &= 0 \end{align}
example
  • q˙\dot{q} is the rate of change of the joint position. (e.g., 0.5 m/s)
  • kk is the unit vector along the axis of translation. The direction of the movement/actuation.

Revolute Joint

v=q˙k×rω=q˙k\begin{align} v &= \dot{q} k \times r \\ \omega &= \dot{q} k \end{align}
  • vv is the velocity at the tool point.
  • ω\omega is the angular velocity of the tool point.
  • q˙\dot{q} is rotation speed. (e.g., 2 rad/s)
  • kk is the unit vector along the axis of rotation. z-axis
  • rr is the length of the link.
example

Velocity for Two Joints

With more than one joint. tool point velocity components are a function of joint velocity and position.

[vω]=f(q˙1,q˙2,q1,q2)\begin{bmatrix} v \\ \omega \end{bmatrix} = f(\dot{q}_1, \dot{q}_2, q_1, q_2)
  • q˙\dot{q} is the rotation speed of the joint.
  • qq is the joint position.

The Jacobian

The Jocobian is a matrix that is a function of joint position, that linearly relates joint velocity to tool point velocity.

[vω]=J(q1,q2)[q˙1q˙2]\begin{bmatrix} v \\ \omega \end{bmatrix} = J(q_1, q_2) \begin{bmatrix} \dot{q}_1 \\ \dot{q}_2 \end{bmatrix}

Elements of the Jocobian

Inverting the Jocobian

To compute the joint velocities for a given tool point velocity, we need to invert the Jacobian.

[vω]=J(q)q˙q˙=J1(q)[vω]\begin{align} \begin{bmatrix} v \\ \omega \end{bmatrix} &= J(q)\dot{q} \\ \dot{q} &= J^{-1}(q) \begin{bmatrix} v \\ \omega \end{bmatrix} \end{align}

The inverse Jocobian can be found by inverting a square portion of the Jacobian.

Find the joint velocities (q˙1,q˙2\dot{q}_1, \dot{q}_2) in terms of the tool point velocity (x˙,y˙\dot{x}, \dot{y}).

Using the Inverse Jocobian

The arm points in the x direction, with joint 2 extended to 0.5 m. Find the joint velocities to move the tool point such that:

x˙=1m/sy˙=1m/s\dot{x} = 1\, \text{m/s} \\ \dot{y} = 1\, \text{m/s}
example
[q˙1q˙2]=1q2[sin(q1)cos(q1)q2cos(q1)q2sin(q1)][x˙y˙][q˙1q˙2]=1q2[010.50][11]=[21]\begin{align} \begin{bmatrix} \dot{q}_1 \\ \dot{q}_2 \end{bmatrix} &= \frac{1}{-q_2} \begin{bmatrix} sin(q_1) & -cos(q_1) \\ -q_2 cos(q_1) & -q_2 sin(q_1) \end{bmatrix} \begin{bmatrix} \dot{x} \\ \dot{y} \end{bmatrix} \\ \begin{bmatrix} \dot{q}_1 \\ \dot{q}_2 \end{bmatrix} &= \frac{1}{-q_2} \begin{bmatrix} 0 & -1 \\ -0.5 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \end{align}
q˙_1=2rad/s;q˙_2=1m/s;\begin{align} \dot{q}\_1 &= 2\, \text{rad/s}; \\ \dot{q}\_2 &= 1\, \text{m/s}; \end{align}

Singularities

A singularity occurs when the Jacobian is not invertible. This happens when the determinant of the Jacobian is zero.

In the previous example, q˙1q2˙\frac{\dot{q}_1}{\dot{q_2}}. If q˙2=0\dot{q}_2 = 0, then the Jacobian is not invertible. There are no valid joint velocity solutions.