How a Trav-A-Dial encoder works

I bought a Trav-A-Dial from a fellow poster here. It's one of the ones with a built in encoder. Southwestern Industries can't or won't tell you what signals it outputs or what voltage it uses. With help from US Digital and from Ned Simmons telling me how to use an oscilloscope to watch the encoder working and google I have been successful in figuring out how to hook up one of these. The operate from 5 volts. On the unit I have the black wire is +5 volts and the white wire is GND. The red and yelow wires are one pair of Nand and And outputs and the violet and green wires are the other pair. The encoder disc has 250 lines on it. Using a digital readout from US Digital, #ED3, configured to read in quadrature, will give the Trav-a-Dial .0001" resolution. From the ED# connect the +5 volts to black, GND to white, channel A to red and channel B to violet. Configure ithe ED3 so that one count equals .0001 and that it counts in quadrature. You will now have a .0001" resolution digital display from your Trav-A-Dial. ERS

Reply to
Eric R Snow
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I don't have much interest in Travadial, but that sounds like a GREAT hacking job you've done there. Congratulations!

DOC

Reply to
DOC

Eric, Did you ever resolve the encoder problem you were fighting with a few weeks ago? If yes, what did it work out to be? Sorry if I missed it at the bottom of one of the threads.

Bob

Reply to
MetalHead

Bob, I'm still working on it. I have made a new pulley with a thread cut into it for the cable to wrap around. I have also made a mirror holder for the laser. But some other more important things had to be done first and I won't be able to get back to it until next week. But I will post the results. Eric

Reply to
Eric R Snow

Thanks Eric, I am curious about that one.

Bob

Reply to
MetalHead

Bob, One of the things I'm not sure about is how to determine the length of an unwrapped helix. If the radius and pitch are known then that should be enough info to calculate the length exactly but I haven't been satisfied with the answers I have come up with. Do you know how to do this? I think that if a right triangle is wrapped around a cylinder then the hypotenuse would be the length of the unwound helix. Do you know if this is correct? Eric

Reply to
Eric R Snow

My college Calc book says that is correct:

for a cylinder of radius A, with a pitch of C, the arc length will be 2 * pi * sqrt(A^2 + C^2)

If you weren't trying for tenths of a mil, I would say ignore the pitch contribution. As it is, probably not. Two other contributers that I don't know the answer to are: cable diameter and wrapping accuracy.

Do you calculate the radius of the pulley at the surface of the pulley, the center of the cable or the outside of the pulley plus the cable?

Wrapping accuracy, does that cable really lay down on the pulley repeatably enough? What kind of tension is required to get that assumption to be valid?

There have been a number of times that I wish I had gone for an Mechanical Engineering degree instead of a Computer Engineering one. As it is, this would be an experimentation project for me. Sorry I can't give you a better answer on that one.

Bob

Reply to
MetalHead

Well, it's good to know I'm on the right track. I don't know, but suspect that the diameter of the cable shouldn't matter. I will be measuring the diameter over the cable when it is under tension to determine where the center of the cable is. With the diameters and units being measured it will be obvious if the cable diameter matters. Think of it this way: If the drum is one inch in circumference and a cable 50 inches in diameter (really limber cable!) is wrapped around it by turning the drum, the free end of the cable would only move 1 inch on the side touching the drum. Right? ERS

Reply to
Eric R Snow

actually the arc length will be:- sqrt((2 * pi *A^2) + C^2)

This is the projection of the hypotenuse formed from the circumference and the helix pitch.

unless I'm very much mistaken.

also the radius should be the radius of the neutral fibre in the cable, which will be the radius for the centre of the cable. This is because the inner side will be compressed to the same extent that the outer side is extended.

HTH Mark Rand RTFM

Reply to
Mark Rand

I have been trying to figure out why the 2pi appears outside the sqrt() also. The calc book is pretty clear on it.

I think that this is going to be one of those try it and see deals, kind of the reason that experienced designers are more valuble than freshouts.

One other source of error that may or may not be a problem is that if the winding and unwinding cable causes the angle between the cable anchor and the wind contact point to change from perpendicular, the encoder will read = sqrt (x^2 + y^2) where Y is the actual separation and X is the moving wind contact point. If the displacement of the wind contact point is less than 1/10 of the minimum separation, this could probably be ignored. Or you could fix it in software.

| X | ===================================== / / / / / / / / | | | | Y | | | O =====================================

It's probably time for me to stop speculating and let Eric build it and see what happens. I am not trying to be negative, I think this is a more complex system that it first appears.

Good Luck, Bob

Reply to
MetalHead

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