Let the forward kinematics map be denoted by $\mathcal{F}$ such that

$\mathcal{F}: \theta \in \mathbb{R}^{n} \rightarrow g \in SE3$

Let the minimal representation of $g$ be given by $x \in \mathbb{R}^{6}$ using axis-angle or other forms of attitude parametrization. If we differentiate the forward kinematics map, we get

$\dot{x} = J_{a}\dot{\theta}$, where $J_{a}$ is the analytical Jacobian. This equation is commonly used in numerical inverse kinematics. However, can we do the reverse?

$x(t_{f})-x(t_0) = \int^{t_{f}}_{t_{0}}J_{a}\dot{\theta}dt$


1 Answer 1



Your equation is true in general only for those manipulators in which $J_a$ is independent of $\theta$ (such as with Cartesian manipulators). Otherwise, the expression is only true in the small (the region of $\theta$ close to $\theta_{t=0}$.

The equation is not true in general because elements of $J_a$ often contain functions of $\theta$. I suppose if you could integrate the expression it would be true, but that just gets you back to your original kinematics function. The Jacobian represents a linearization of the kinematic mapping. It is only a first-order approximation, so the expression does not hold globally.


If you do want to numerically integrate using your equation, you would need to modify (update) $J_a$ at each iteration based on the new values of $\theta$. This would then be similar to the algorithm for velocity control, although I believe your timestep would be finer.

  • $\begingroup$ For a time parametrized path in joint space $\theta(t), \dot{\theta}(t) \forall t \in [t_{0}, t_{f}]$, would a numerical integration make sense, like $x(t_{f}-x(t_{0}) = \sum^{N}_{k=0}J_{a}(\theta_{k})\dot{\theta_{k}}\Delta t$ where $k$ is the discrete sample time and $\Delta t$ is the step size. $\endgroup$
    – jjgarrison
    Dec 18, 2016 at 23:30
  • $\begingroup$ I think it does if you update the values of $J_a$ at each iteration. You may want to research velocity control, which this comes pretty close to. $\endgroup$
    – SteveO
    Dec 18, 2016 at 23:35
  • $\begingroup$ Why wouldn't this be valid for Cartesian manipulators? $\endgroup$
    – 50k4
    Dec 19, 2016 at 5:57
  • $\begingroup$ Never mind, I understood what you meant. For Cartesian it is generally valid, no need to resample for every time instance. $\endgroup$
    – 50k4
    Dec 19, 2016 at 5:58
  • $\begingroup$ Edited to make information about Cartesian manipulators more clear and to include numerical integration answer which was in the comments. $\endgroup$
    – SteveO
    Dec 19, 2016 at 14:29

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