In this paper by J. W. Burdick, the main result basically says that

for a redundant manipulator, infinity solutions corresponding to one end-effector pose can be grouped into a finite set of smooth manifolds.

But later in the paper, the author said only revolute jointed manipulators would be considered in the paper.

Does this result (grouping of solutions into a finite set of manifolds) hold for redundant robots with prismatic joint(s) as well? Is there any significant difference in analysis and result when prismatic joints are included? So far I couldn't find anyone explicitly address the case of robots with prismatic joints yet.

(I am not sure if this site or math.stackexchange.com would be the more appropriate place to post this question, though.)


2 Answers 2


Saw this question and figured I would throw my two cents in 5 years later.

Gilhad's reasoning about the mathematical process of simple to complex theorems/proofs makes sense as to why the analysis is done only with revolute joint manipulators. However, going from spherical to prismatic is going from complex to simple. A solely prismatic serial manipulator is painfully easy to analyze:

Consider the fact that a prismatic joint has no rotational axis. Thus, a solely prismatic serial manipulator has no rotational degrees of freedom. This is why there is no real discussion about such manipulators - they are a trivial case (constrained to only 1D, 2D, or 3D translations). The aforementioned transformation from revolute joints to prismatic joint would suffer from the same issues - as it has been specifically designed to be "translate-only".

The analysis of serial manipulators becomes much more difficult when revolute joints are introduced - bringing with them lots of sine's and cosine's, as well as (up to) 3 additional orientation degrees of freedom.

As for what the analysis of a prismatic joint serial manipulator looks like, let's consider our end-effector Jacobian at a given joint configuration $q$, $J(q)$ - the crystal ball of manipulator properties. Given a robotic manipulator with $n$ joints (with some trivial design constraints being satisfied) and $m$ degrees-of-freedom in the workspace, Burdwick's analysis is restricted to the cases where $n \ge m$. Thus, $J \in \mathbb{R}^{m\times n}$ is either a square matrix or a matrix such that there is at least one linearly dependent column (this latter idea being useful for the derivation of the aforementioned smooth manifolds).

To gain some intuition on a general prismatic manipulator Jacobian, let's look at a 3-joint prismatic serial manipulator where joint one translates the end-effector solely along the X-axis, joint 2 along the Y-axis, and joint 3 along the Z-axis. We will assume this analysis is being done in the future where problems do not exist, such that the manipulator has no discernible joint limits. Without getting into the derivation of this manipulator's Jacobian, we will wave our hands and find that: $$J(q) = \left[ {\begin{array}{cc} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} } \right]$$ We see that $J(q)$ is invertible, a necessary condition for a prismatic manipulator to have 3 workspace degrees-of-freedom, for all possible joint configurations. Thus, any movement in our workspace, defined as $\dot{x} = J(q)\dot{q}$, can be directly mapped back to a movement in the joint space as $\dot{q} = J(q)^{-1}\dot{x}$. Intuitively, we can see that following a desired joint trajectory will cause a unique end-effector motion - and likewise a unique end-effector trajectory will be caused by a unique joint trajectory (not at all the case with revolute joint manipulators). With a few additional steps we could prove that the forward kinematics map of this manipulator, and in fact all prismatic manipulators where $n = m \le 3$ and $J(q)$ is invertible, is a bijection.

This is a distinct difference from one of the major propositions Burdick makes about revolute manipulators - which is the forward kinematics map for a 3-joint revolute spatial (having all 3 translational degrees-of-freedom) manipulator may have up to 4 unique solutions.

Finally, when considering redundant manipulators - one interesting addition to the original problem of inverse kinematics is that each solution will lie on a smooth $n-m$ dimensional manifold (as opposed to a single point). To traverse this manifold, we begin by taking the singular value decomposition of $J(q)$. This yields the following relationship: $J(q) = U\Sigma V^{T}$, where $U \in \mathbb{R}^{m\times m}$, $\Sigma \in \mathbb{R}^{m\times n}$, $V \in \mathbb{R}^{n\times n}$, $U$ and $V$ are orthonormal matrices, and $\Sigma$ has non-zero entries only along its diagonal. As the last $n-m$ columns of $\Sigma$ contain only zeroes, the last $n-m$ rows in $V^{T}$ will serve as a basis for "null" motion (joint motion that does not affect the end-effector). If we let $v = \dot{q}$ be equal to one of these vectors (or a linear combination of them), then $V^{T}v \in \mathbb{R}^{n\times 1}$ will have non-zero entries only in the last $n-m$ rows (perhaps an interesting exercise for the curious). These non-zero entries will be "zeroed" by the $n-m$ all-zero columns of $\Sigma$. This should be enough to show that we satisfied the following: $$ \dot{x} = J(q)\dot{q} = U\Sigma V^{T}\dot{q} = 0, \dot{q} = v$$

Given that a prismatic joint manipulator has a constant Jacobian, its singular value decomposition will also be constant and thus linear. The manifolds for these prismatic manipulators then are $n-m$ dimensional linear subspaces of the $n$ dimensional joint space.

There, I wasted my time explaining the analysis of prismatic joint manipulators to show why you will never see anyone else wasting their time on the analysis of prismatic joint manipulators ;)


I am not sure if it applies here, but it is usual way in mathematic to prove something simple and then say, that other cases can be transformed to the simple case, so it is proven for them too (usually in some ineffective long way, but it does not matter, if you want to prove existence of solution).

Prismatic joint can be replaced with three revolute joints connected with long enough rodes, working in cooperative way, maybe it is enought to do substitution for the case in the papers?

  • $\begingroup$ Hm, that's interesting. I never realize before that a prismatic joint can be replaced by three revolute joints with some constraints. That might be a way to go. $\endgroup$ Commented Mar 29, 2017 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.