I am "working" on having a motorized feed for my lathe. The VFD control
on my drill press is SO convenient that I think it would be nice on the
lathe too. The biggest problem is getting the speed range, but I also
needed to know how much torque is required, for motor sizing.

I have already extended the lead screw on the tailstock end, so I made
an adapter that fits over that and has a 3/8" diam stub. I chucked the
stub in a cordless drill, set the DOC worst case, started the lathe,
then started the feed with the drill. As it was cutting I turned down
the clutch on the cordless drill until it slipped. That setting being
too small, I increased it one step.

I then took the drill and chucked up an L-shaped rod, with a 10" free
leg. Putting the end of that leg on a scale, I very slowly pulled the
trigger and noted the scale at the point the clutch slipped. It was 1
lb - 8 oz. Times the 10" arm gives 240 in-oz, or 15 in-lb torque.
Surprisingly small!

Admittedly, this is not a very precise way to measure torque, but it's
close enough for me. I'll probably double it when I come to chose the
motor needed, just to be conservative. YMMV.

Bob

Bob Engelhardt writes:

Admittedly, this is not a very precise way to measure torque, but it's
close enough for me.

Drill improvised into a dynamometer. Not bad.

I would consider a PMDC motor and speed controller instead of AC and VFD.
Better economics in this low-power size, and a gutsy low-speed torque
reserve.

XL or L belt and pulleys will solve the shaft alignment problem, while
allowing you to change the ratio if your torque target turns out wrong.
Will avoid slippage that will make your feed (and part finish) uneven.

"Bob Engelhardt" wrote in message
. ..
I am "working" on having a motorized feed for my lathe. The VFD control
on my drill press is SO convenient that I think it would be nice on the
lathe too. The biggest problem is getting the speed range, but I also
needed to know how much torque is required, for motor sizing.

I have already extended the lead screw on the tailstock end, so I made
an adapter that fits over that and has a 3/8" diam stub. I chucked the
stub in a cordless drill, set the DOC worst case, started the lathe,
then started the feed with the drill. As it was cutting I turned down
the clutch on the cordless drill until it slipped. That setting being
too small, I increased it one step.

I then took the drill and chucked up an L-shaped rod, with a 10" free
leg. Putting the end of that leg on a scale, I very slowly pulled the
trigger and noted the scale at the point the clutch slipped. It was 1
lb - 8 oz. Times the 10" arm gives 240 in-oz, or 15 in-lb torque.
Surprisingly small!

Admittedly, this is not a very precise way to measure torque, but it's
close enough for me. I'll probably double it when I come to chose the
motor needed, just to be conservative. YMMV.

Bob

Am I missing something here???

It doesn't take much torque to simply turn the leadscrew (and, presumably,
move the carriage. But, when you are trying to take a heavy cut, the torque
can be pretty substantial...

Jerry

Jerry Foster wrote:
Am I missing something here???

It doesn't take much torque to simply turn the leadscrew (and, presumably,
move the carriage. But, when you are trying to take a heavy cut, the torque
can be pretty substantial...

I think that you did miss it. I can be too brief in my posts, so it was
probably easily missed: I took the torque measurement during a heavy
cut, the heaviest that I would expect to make (.050" DOC & .050 advance,
IIRC - it's not a heavy lathe).

Bob

I think that you did miss it. I can be too brief in my posts, so it was
probably easily missed: I took the torque measurement during a heavy cut,
the heaviest that I would expect to make (.050" DOC & .050 advance, IIRC -
it's not a heavy lathe).

Bob

for what it's worth, the manual for my logan PL92 lathe says that the clutch
slips at 900 pounds force



--
Posted via a free Usenet account from http://www.teranews.com

On Jan 12, 12:11 pm, Bob Engelhardt wrote:
I am "working" on having a motorized feed for my lathe. The VFD control
on my drill press is SO convenient that I think it would be nice on the
lathe too. The biggest problem is getting the speed range, but I also
needed to know how much torque is required, for motor sizing.

The manual for my Atlas-Craftsman 12x36 cabinet lathe (underdrive)
says to set the leadscrew safety clutch at 5 ft-lbs. It has only
kicked in once, when a chip worked its way into the apron gearing and
jammed the entire mechanism. I generally take fairly heavy cuts to
maximize productivity.

Richard J Kinch wrote:
....
I would consider a PMDC motor and speed controller ...

I have been considering that combo. But I'm looking for about 30 rpm on
the lead screw and I don't know if that's doable. My short list
includes a cordless drill motor & gearbox.

XL or L belt and pulleys will solve the shaft alignment problem, ...

That's something new. How do they do that?

Will avoid slippage that will make your feed (and part finish) uneven.

Good point.

Thanks,
Bob

William Noble wrote:
for what it's worth, the manual for my logan PL92 lathe says that the clutch
slips at 900 pounds force

YOWEE! Are you sure? There's a unit missing there, but even if it's
900 INCH-pounds, that's a LOT of torque (75 ft-lbs)!

Or maybe that's the axial force on the clutch plate. In which case it
really doesn't say anything about the torque.

Bob

woodworker88 wrote:
The manual for my Atlas-Craftsman 12x36 cabinet lathe (underdrive)
says to set the leadscrew safety clutch at 5 ft-lbs. It has only
kicked in once, when a chip worked its way into the apron gearing and
jammed the entire mechanism. I generally take fairly heavy cuts to
maximize productivity.

That's really useful information, thanks! If 5 ft-lbs is the safety
limit on a 12x36, the 1 1/4 ft-lbs (15 in-lbs) that I measured for a
heavy cut on my 10x24 sounds realistic.

Thanks,
Bob

Bob Engelhardt writes:

XL or L belt and pulleys will solve the shaft alignment problem, ...

That's something new. How do they do that?

How were you planning to couple the motor to the leadscrew? You must have
some sort of coupling that transmits torque through misaligned shafts.
Belt-and-pulleys is one method. For direct methods:

http://groups.google.com/group/rec.c...f26dc704d6d31c

On Sun, 13 Jan 2008 20:20:49 -0500, Bob Engelhardt
wrote:

William Noble wrote:
for what it's worth, the manual for my logan PL92 lathe says that the clutch
slips at 900 pounds force

YOWEE! Are you sure? There's a unit missing there, but even if it's
900 INCH-pounds, that's a LOT of torque (75 ft-lbs)!

Or maybe that's the axial force on the clutch plate. In which case it
really doesn't say anything about the torque.

Bob

Actually, it's 900 pounds of THRUST on the carriage.


--
+--------------------------------------------+
| Scott Logan - ssl "at" lathe.com |
| Logan Actuator Co. http://www.lathe.com |
| Harvard, IL |
|++++++++++++++++++++++++++++++++++++++++++++|
| Parts and Accessories for Logan Lathes and |
| Montgomery Wards Lathes |
| Logan-Lilly Mine Hoist Safety Controllers |
+--------------------------------------------+
"Measure Twice, Cut Once"

RCM FAQ - http://w3.uwyo.edu/~metal
Metal Web News - http://www.metalwebnews.com/
Help squash SPAM: http://www.cauce.org/

Scott S. Logan wrote:
Actually, it's 900 pounds of THRUST on the carriage.

900 lbs! So, if you rigged some pulleys, the lathe could lift itself
off the ground? That might be convenient for getting it onto skates for
moving BG

As far as lead screw torque: an 1/8" pitch, 3/4" diameter leadscrew has
a PI*3/4 / 1/8 mechanical advantage. About 19:1. So 900 lbs thrust
means 900/19 = 47 lbs tangentially on the screw. 47 * 3/8 (radius) = 17
in-lbs torque. I've ignored the friction between the leadscrew and its
half nut, as Mach. Handbook gives a value of .04 for worm gear at 250 fpm.

I don't know how big a "PL92" lathe is, but a clutch slip point of 17
in-lbs seems pretty low. The slip point is surely well above the normal
operating force. I measured 15 in-lbs on my leadscrew, but there is no
claim of precision or accuracy. Also, my assumption about the leadscrew
pitch could be wrong. The larger the pitch, the larger the torque for
the same thrust. (The assumption about the leadscrew diameter is
irrelevant - the diameter cancels out in the calculations.) If the
pitch were 1/4", the torque required to generate 900 lbs thrust would be
34 in-lbs. Or, my reasoning could be all wrong!


Bob

On Mon, 14 Jan 2008 11:11:00 -0500, Bob Engelhardt
wrote:

Scott S. Logan wrote:
Actually, it's 900 pounds of THRUST on the carriage.

900 lbs! So, if you rigged some pulleys, the lathe could lift itself
off the ground? That might be convenient for getting it onto skates for
moving BG

As far as lead screw torque: an 1/8" pitch, 3/4" diameter leadscrew has
a PI*3/4 / 1/8 mechanical advantage. About 19:1. So 900 lbs thrust
means 900/19 = 47 lbs tangentially on the screw. 47 * 3/8 (radius) = 17
in-lbs torque. I've ignored the friction between the leadscrew and its
half nut, as Mach. Handbook gives a value of .04 for worm gear at 250 fpm.

You can't ignore friction. The efficiency of a 3/4-8 acme screw is
perhaps 30% - a consequence of the fact that the load is carried near
the OD of the screw.

As you've already figured, the OD of your screw travels 19x the
distance the load moves, and the effect of the friction is multiplied
as well - 1-(19 x .04) = .24 efficiency

Assuming the load is actually carried at the pitch diameter, and
guesstimating the the PD of your screw as .68 inch -
1-(17 x .04) = .32 efficiency

That's a grossly simplified analysis, but it works reasonably well for
acme screws, less well for 60 degree threads.

--
Ned Simmons

Ned Simmons wrote:
You can't ignore friction. ...

... = .24 efficiency

.....

Thanks. I don't follow, but you have the credibility for me to accept
it nevertheless.

Assuming "efficiency" relates to force in this case, the frictionless 17
in-lbs that I calculated really requires 17/.24 = 71 in-lbs (6 ft-lbs).
For clutch release, which would be well above the normal operating
force. Which is more consistent with my 15 in-lbs (1 1/4 ft-lbs)
operating torque measure. Also right in line with the 5 ft-lbs that
"woodworker88" quoted for his 12x36.

Here's mo
http://www.racointernational.com/screw_formula.htm

It's a calculator for lead screw parameters. Giving it 900 lbs thrust,
..68 PD, .125 lead/pitch, and .04 COF returns 30 in-lbs. It doesn't say
what the model is.

Thanks,
Bob

On Mon, 14 Jan 2008 13:29:31 -0500, Bob Engelhardt
wrote:

Ned Simmons wrote:
You can't ignore friction. ...

... = .24 efficiency

.....

Thanks. I don't follow, but you have the credibility for me to accept
it nevertheless.

It helps if you think of the screw as an inclined plane wrapped
wrapped around the pitch diameter of the screw. Now imagine pushing a
weight up the incline. In order to raise the weight one unit you need
to slide it (while overcoming friction) a much greater distance (17x
in this case).

And now that I think about it that way, I see that the approximation
of efficiency I wrote earlier is even sloppier than I thought. This
should be better:

(Lead / (pi * PitchDia)) / ((Lead / (pi * PitchDia)) + CoeffFriction)

This looks like a similar approach.
http://www.utm.edu/departments/engin...Notes%2027.pdf

This is the more complete, but not very intuitive analysis you'll find
in machine design texts.
http://www.roymech.co.uk/Useful_Tabl..._Screws_1.html

--
Ned Simmons

"Bob Engelhardt" wrote in message
...
Scott S. Logan wrote:
Actually, it's 900 pounds of THRUST on the carriage.

900 lbs! So, if you rigged some pulleys, the lathe could lift itself off
the ground? That might be convenient for getting it onto skates for
moving BG

As far as lead screw torque: an 1/8" pitch, 3/4" diameter leadscrew has a
PI*3/4 / 1/8 mechanical advantage. About 19:1. So 900 lbs thrust means
900/19 = 47 lbs tangentially on the screw. 47 * 3/8 (radius) = 17 in-lbs
torque. I've ignored the friction between the leadscrew and its half nut,
as Mach. Handbook gives a value of .04 for worm gear at 250 fpm.

I don't know how big a "PL92" lathe is, but a clutch slip point of 17
in-lbs seems pretty low. The slip point is surely well above the normal
operating force. I measured 15 in-lbs on my leadscrew, but there is no
claim of precision or accuracy. Also, my assumption about the leadscrew
pitch could be wrong. The larger the pitch, the larger the torque for the
same thrust. (The assumption about the leadscrew diameter is irrelevant -
the diameter cancels out in the calculations.) If the pitch were 1/4",
the torque required to generate 900 lbs thrust would be 34 in-lbs. Or, my
reasoning could be all wrong!


Bob


1. the lathe weighs closer to 1200 pounds so it probably couldn't lift
itself, though I'm sure Scott could provide the exact weight if the urge
struck him

2. I think you are off by a factor of 12 (feet to inches) - specifically
it's 47*12*3/8 =
or alternatively, you meant 17 foot pounds, not inch pounds.



--
Posted via a free Usenet account from http://www.teranews.com

Bob Engelhardt wrote:

It's a calculator for lead screw parameters. Giving it 900 lbs thrust,
.68 PD, .125 lead/pitch, and .04 COF returns 30 in-lbs. It doesn't say
what the model is.

Where did you get that figure of .04 for what I assume is brass on steel?

Wes