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

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

DOC


"Eric R Snow" wrote in message
...
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

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

On Sun, 30 Oct 2005 15:38:06 GMT, MetalHead
wrote:

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
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

Eric R Snow wrote:
On Sun, 30 Oct 2005 15:38:06 GMT, MetalHead
wrote:


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

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

Thanks Eric, I am curious about that one.

Bob

On Sun, 30 Oct 2005 19:23:31 GMT, MetalHead
wrote:

Eric R Snow wrote:
On Sun, 30 Oct 2005 15:38:06 GMT, MetalHead
wrote:


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

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

Thanks Eric, I am curious about that one.

Bob
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

Eric R Snow wrote:
On Sun, 30 Oct 2005 19:23:31 GMT, MetalHead
wrote:


Eric R Snow wrote:

On Sun, 30 Oct 2005 15:38:06 GMT, MetalHead
wrote:



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

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

Thanks Eric, I am curious about that one.

Bob

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

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 a 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

On Mon, 31 Oct 2005 00:24:21 GMT, MetalHead
wrote:

Eric R Snow wrote:
On Sun, 30 Oct 2005 19:23:31 GMT, MetalHead
wrote:


Eric R Snow wrote:

On Sun, 30 Oct 2005 15:38:06 GMT, MetalHead
wrote:



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

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

Thanks Eric, I am curious about that one.

Bob

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

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 a 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

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

On Mon, 31 Oct 2005 05:36:40 GMT, (Eric R Snow) wrote:

On Mon, 31 Oct 2005 00:24:21 GMT, MetalHead
wrote:

Eric R Snow wrote:
On Sun, 30 Oct 2005 19:23:31 GMT, MetalHead
wrote:


Eric R Snow wrote:

On Sun, 30 Oct 2005 15:38:06 GMT, MetalHead
wrote:



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

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

Thanks Eric, I am curious about that one.

Bob

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

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 a 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

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


pedant mode

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.
/pedant mode

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

Mark Rand wrote:
On Mon, 31 Oct 2005 05:36:40 GMT, (Eric R Snow) wrote:


On Mon, 31 Oct 2005 00:24:21 GMT, MetalHead
wrote:


Eric R Snow wrote:

On Sun, 30 Oct 2005 19:23:31 GMT, MetalHead
wrote:



Eric R Snow wrote:


On Sun, 30 Oct 2005 15:38:06 GMT, MetalHead
wrote:




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

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

Thanks Eric, I am curious about that one.

Bob

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

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 a 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


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



pedant mode

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.

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


unless I'm very much mistaken.
/pedant mode

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.

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