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I have only recently started reading Steve LaValle's book on motion planning. It mentions that a rotating link with a fixed pivot has a configuration space of a unit circle.

Since the configuration space is the set of all possible configurations the link can have, i.e. all possible angles from 0° to 360°, shouldn't the c-space be a line rather than a circle?

Also, for a 2 link robotic arm in the same plane, how is the c-space a 2-tourus even though the joints rotate in the same plane?

Lastly what's the difference between the representations shown below? enter image description here

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Since the configuration space is the set of all possible configurations the link can have, i.e. all possible angles from 0° to 360°, shouldn't the c-space be a line rather than a circle?

You are correct that a rotating link has one variable $\theta$ that can attain any value in the range $[0, 2\pi]$ (let's talk in radian). However, the configuration space of the link is not simply a line segment because the link being at $\theta = 2\pi$ is identical to it being at $\theta = 0$. In other words, the link can keep rotating forever while the variable $\theta$ simply runs in loops, like a clock.

If there was an ant walking on the line segment to the right, when the ant reaches $\theta = 2\pi$, it would somehow be teleported and appear at $\theta = 0$ and continue walking like nothing happened.

This is essentially the same as gluing both ends of the line segment together. That's why you have a circle instead of only a line segment.

Also, for a 2 link robotic arm in the same plane, how is the c-space a 2-tourus even though the joints rotate in the same plane?

I understand that this may be confusion between the robot's workspace and the robot's configuration space.

The joints rotate in the same plane so their workspace is a subset of that two-dimensional plane.

Now for the configuration space of this robot, the same argument as above applies. Initially, you can think of a configuration space as a square whose corners are $(\theta_1, \theta_2) = (0, 0)$, $(\theta_1, \theta_2) = (0, 2\pi)$, $(\theta_1, \theta_2) = (2\pi, 2\pi)$, and $(\theta_1, \theta_2) = (2\pi, 0)$.

The first link (the one attached to the base) of the robot could continue rotating forever while the joint value $\theta_1$ would only vary within the range $[0, 2\pi]$. If there was an ant walking from left to right on the square, the ant could continue walking forever as previously. It is like the left edge and the right edge are glued together.

The second link could also keep rotating forever while the joint value $\theta_2$ could only run between $0$ and $2\pi$. In the end, this is like gluing the top edge and the bottom edge together.

In topology, this gluing process is called identification.

Finally, when you glue the two pairs of edges of the square together, you get a 2-torus.

torus from rectangle

(image from https://en.wikipedia.org/wiki/Torus)

Lastly what's the difference between the representations shown below?

The two images show the same thing before and after the identification (gluing) process. Notice the arrow heads on the edges of the square in the left image. They tell you which sides match and how they should be aligned when being glued together.

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  • $\begingroup$ Brilliant! Thanks a lot. But now if joint limits where to be introduced, the 'polygon' won't be glued together, isn't it? Hence a n-dimensional arm will always have a n-dimensional c-space as long as there are joint limits? $\endgroup$ – harsh Oct 17 '17 at 7:17
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    $\begingroup$ Actually the configuration space would remain the same. What changes is the free subset of c-space. Now you have introduced some obstacles that potentially prevent the ant from walking indefinitely from left to right. And an $n$-jointed arm will always have an $n$-dimensional c-space, independent of the joint limits. $\endgroup$ – Petch Puttichai Oct 17 '17 at 7:27
  • $\begingroup$ And an n-jointed arm will always have an n-dimensional c-space, independent of the joint limits. But in my example, a 2D arm has a 3D (tourus) c-space. $\endgroup$ – harsh Oct 17 '17 at 7:29
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    $\begingroup$ You have hit the point! The configuration space is that torus. But a torus is the surface of that doughnut. It is essentially a 2-dimensional manifold that lives in 3-dimensional space (that's why it's called a 2-torus). The concept of manifolds is pretty complicated and very confusing for me at first. Roughly speaking, a manifold is a smooth surface that lives in a higher-dimensional space. $\endgroup$ – Petch Puttichai Oct 17 '17 at 7:35
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    $\begingroup$ I was, too :) It just takes time to really digest the content. I'd say I'm far from truly understanding the topic but if you want some words: basic topology books will definitely help if you really want to look into the topic. Otherwise, I think that just from reading many papers (in motion planning, etc.) where people talk about manifolds differently, you will gradually get the clearer picture. $\endgroup$ – Petch Puttichai Oct 17 '17 at 7:53

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