8
$\begingroup$

I'm building a motion control platform with 3 DoF: 1 axis of rotation (theta) and 2 cartesian (x,y). In most applications, like wrist actuation, you have an X-Y stage with a rotating servo as the stage's payload. This configuration works well since little of the power and data wiring needs to transit to the non-linear moving portion of the platform.

For my inverted application, the stackup is reversed. The rotating axis comes first (from the mounting plane) with the stage connected as the rotating platform's payload. Now nearly all of the wiring (power, command, sensor, and otherwise) must be routed to the non-linearly moving section.

I can see two broad approaches:

  1. The inside track, I route the cabling through the center of rotation.

  2. The outside track, I route the cabling around outside the outer diameter of the rotating platform.

Mathematically, I can see that (1) results in minimum cable length, but maximum torsional loading, while (2) results in maximum cable length, but minimum torsional loading on the wires.

Having limited experience with cable routing (and the associated carriers, strategies, and products) in non-linear applications, my question is...

...which approach is better in practice?

Cost isn't really the issue here. I'm more interested in reliability, ease of construction, availability of commercial components (says something about the popularity of the technique), etc...

e.g. the generic concepts behind why you pick one over the other.

...of course, if you have some part numbers for me I wouldn't be upset <-- I know I'm not supposed ask that here ;-)

$\endgroup$
  • 1
    $\begingroup$ Interesting question, I've stumbled across this problem quite a few times. Frustrating, but it's a learning experience :) I'll see if I can collect my thoughts into a good answer later. $\endgroup$ – Manishearth Dec 16 '12 at 5:30
  • $\begingroup$ I don't think the route is what affects cable torsion - rather the slack in the cables will determine how much torsional force is applied to the cables - the option probably depends on your setup, as to which is easier to accomplish (space available vs number of cables and whether it will fit) $\endgroup$ – ronalchn Dec 16 '12 at 5:32
  • 1
    $\begingroup$ Either way, make sure you buy cable that is specifically designed for "high-flex" operations, this will last much longer. $\endgroup$ – WildCrustacean Dec 16 '12 at 13:07
  • $\begingroup$ This is neither an answer nor a criticism, but if you have any diagram of your X/Y/$\theta$ setup then please add it. Even though the core of your question is about cabling, being able to better visualize the problem will add more context to the solution. $\endgroup$ – Ian Dec 18 '12 at 22:45
  • $\begingroup$ Thanks. I really wanted to. I'm working on a model. I'll get it uploaded as soon as I can. Apologies for the delay. $\endgroup$ – DrFriedParts Dec 19 '12 at 8:43
2
$\begingroup$

I've seen a number of systems in this configuration and most went for an outside track solution. Part of the reason for this is control of bend radius. With an outside track, the bend is obvious at all positions and it is clear when you run out of track.

If you are bothered about cabling complexity, you could put more of the electronics on the rotated stage, so instead of having motor, encoder and other cables all running down the energy chain, you would just have power and data lines, with everything else done by remote i/o.

Taking this to the extreme, I've worked at a place where this technique was used with slip rings for a continuously revolving robot. It had two scara arms and all of the control electronics for them mounted on a revolving platform. The data slip rings were horribly noisy, so the data connection had to have more ECC than normal, but it all worked well.

$\endgroup$
  • $\begingroup$ The other problem with slip rings is that they introduce small changes in resistance that can cause the capacitors in your motor controller to fail. $\endgroup$ – Ian Dec 22 '12 at 2:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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