Inverse kinematics and the Jacobian transpose


This is a relatively technical post. Its purpose is mainly to teach myself why the Jacobian transpose is so useful when doing inverse kinematics.

We are going to solve the following problem:

We have a mechanism with effectors applying forces whose components are given by f_i, with i going from 1 to N. The required power is provided by a series of torques, whose components are called \tau_j, where j goes from 1 to M. Get the required values of \tau_j as a function of f_i.


Let’s call \theta_j the angular coordinate associated with the torque components \tau_j and x_i the normal (“linear”) coordinate of the effector associated with the force component f_i. In most useful cases, the positions of the effectors can be expressed as a function of the angular coordinates, x_i(\theta_j). The Jacobian will be the linearization of this relationship around some point \theta_j^0,

\displaystyle J_{ij}(\theta_j^0) = \left. \frac{\partial x_i}{\partial \theta_j} \right|_{\theta_j=\theta_j^0}.

Virtual work

If we can ignore inertia forces (either because we are dealing with a purely static problem or because inertia is negligible), we can get

\displaystyle \sum_{i=1}^N f_i \delta x_i = \sum_{j=1}^M \tau_j \delta \theta_j,

where \delta x_i is the infinitesimal linear displacement associated with the force component f_i and \delta \theta_j is the infinitesimal angular displacement associated with the torque component \tau_j.

Putting things together

As the previous expression uses infinitesimal movements, we can use the Jacobian to relate the linear displacements to the angular ones:

\displaystyle \delta x_i = \sum_{j=1}^M J_{ij} \delta \theta_j.

If we replace that result in the virtual work equation,

\displaystyle \sum_{i=1}^N f_i \left(\sum_{j=1}^M J_{ij} \delta \theta_j\right) = \sum_{j=1}^M \tau_j \delta \theta_j,

and we do some rearrangements, we get an expression with infinitesimal angular displacements in both sides:

\displaystyle  \sum_{j=1}^M \left(\sum_{i=1}^N f_i J_{ij}\right) \delta \theta_j = \sum_{j=1}^M \tau_j \delta \theta_j.

As the infinitesimal angular displacements \delta \theta_j are arbitrary, their factors should match:

\displaystyle \sum_{i=1}^N f_i J_{ij} = \tau_j.

By representing this equation in matrix form,

\displaystyle \boldsymbol{\tau} = \mathbf{J}^T \mathbf{f},

we see how we arrive naturally to the Jacobian transpose.


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