Re: [CAD_CAM_EDM_DRO] where's the torque ?

Les,
As you know, and most of the list doesn't, I'm more of an empirical
engineer, first theory, then practical. In my experience the ratios I
quoted get you pretty close to where you need to be to get decent
performance, but like everything else in life, if you want that alst 10%
or so, it gets expensive quickly. About 80% of my CNC work is retrofits,
and those numbers have gotten me in the ballpark every time. After the
first few machines, driving myself (and my loving wife) mad with a
calculator and a scratch pad, I kept getting approximately the same
numbers. They work and quite well, for 90% of the machines out there. I
use my monster machine as a test bed, and have played more with drive
systems in the last 2 years than most do in a lifetime, the present
iteration has quite a bit more reduction than I prefer for steppers,
though it was a trade off between accel. (actually CV path accuracy) and
speed. I do have a set of servos to bolt on, and my do that someday, as
I'm looking at machining some PVC and could use the higher feed rates.


erie

Leslie Watts wrote:

>Unlike leadscrews I think rack and timing belt drive will need
>a lot of motor reduction. I think Erie's figures are pretty
>close to the mark. I'll do a run on Motioneering and see
>how it looks.
>
>Les
>
>Leslie M.Watts
>L M Watts Furniture
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>-----Original Message-----
>From: CAD_CAM_EDM_DRO@yahoogroups.com
>[mailto:CAD_CAM_EDM_DRO@yahoogroups.com]On Behalf Of R Rogers
>Sent: Monday, May 16, 2005 1:07 PM
>To: CAD_CAM_EDM_DRO@yahoogroups.com
>Subject: Re: [CAD_CAM_EDM_DRO] where's the torque ?
>
>
>
>
>turbulatordude <dave_mucha@...> wrote:
>I've been running thru the torque numbers for motors and seem to be
>missing something.
>
>A 1-1/2hp treadmill motor with 5,100 RPM has 296 oz-in torque.
>
>On a rack and pinion. and using a 1.25" pitch, we can round the
>distance per gear rotation to 4 inches. 1.25*pi=3.926, so 4 is close
>enough.
>
>When we calculate for a leadscrew, we come back to inches of table
>movement. do we also do that for routers ?
>
>I mean that we put on a 600oz-in motor, then multiply that by 5TPI to
>get 3,000 OZ-IN that would mean 3,000 per inch, no ?
>
>/// Yeah, if we were compressing a spring and wanted to calculate the
>kinetic force over distance. Distance traveled has nothing to do with
>torque. Torque, speed and power are determined by the size of the motor and
>the reduction being used. The pulsetrain determines the distance.///
>
>That would make the motor 296/4=74 oz-in if it were direct drive.
>using 6:1 pulleys would then multiply that up to 444 oz-in. or 444 per
>inch ?
>
>The leadscrew moves 1 inch per 5 motor rev or 0.2" per rev while the
>rack moves 1.5 revs per in. Gearing up to get 5 rev's per inch would
>mean a 20:1 pulley to the 1:4 rack gear. If course that would
>multiply the 296 by 5 for 1,482oz-in.
>
>Since I've seen routers and plasma cutter run with much smaller motors
>and much smaller ratios (3:1 on a 350 oz-in stepper) it seems I'm
>missing something.
>
>///Steppers can get away with this because they like operating near their
>holding torque, very low RPM. I'd think if one designs a stepper system that
>never exceeded a few hundred RPM it would perform well. Servos on the other
>hand do not operate well at low speeds and need alot of reduction to rapid
>at 80% of their top speed.
>
>The bottom line is: If you plan to use servos with a rack and pinion
>arrangement, you are going to need alot of reduction to do so. Example: 3000
>RPM servo and 1:4 pinion to rack ratio with 100 ipm rapids. 80% of 3000=
>2400 RPM. 100ipm/4=25 RPM of the pinion gear. 2400/25=96. With this config
>you'd need a 96:1 ratio. /// Ron
>
>Anyone see my mistakes or what I've overlooked ?
>
>Dave
>
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