[CAD_CAM_EDM_DRO] treadmil motor speed reduction analysis
Posted by
Leslie Watts
on 2005-05-17 10:22:55 UTC
Hello all,
As promised I did an analysis of a hypothetical cnc router
with belt or rack drive using treadmill motors.
I had to do some eductated guesses...because there are few
specs on them. These are the cheap 1-2 hp 100vdc units
at surplus center and others.
First, I am going to seriously derate these things because they are
"treadmill" duty. I don't believe the HP ratings anyway. Let's derate
them 50%.
Second I have to get some idea of the rotor moment of inertia. None have
such specs, but I know a 3/4 hp square frame typical PMDC motor comes
out at about 0.4 oz*in*sec^2. I will use that figure. This is a very
high inertia compared to a typical industrial servo and will be one of the
biggest factors limiting performance.
Here is the router specs:
100 kG moving mass
125N (25 lb) cutting force
50 N (10 lb) friction force
100 mm/sec (4 inches/sec) max speed
.25g (2.45 M/sec^2) max acceleration
50mm (2 in) rack pinion/timing pulley diameter
Ok.
First the moment of inertia at the pulley/pinion.
That is simply Mr^2= 100*(.025M^2)= 8.8 oz*in*sec^2
A 1:1 load to motor inertia match occurs with SQRT( 8.8/.4)=
4.69 gear reduction ratio. Lets make it an even 5:1.
Starting with that I went to motioneering and
checked it with real motors.
I needed 600 in oz peak at only 171 rpm.
Peak input power was only 90 watts.
But....this is a no go because I have derated the motor.
I don't think they can put out much more than 400 in oz peak
and much less continuous.
Let's go with 10:1 gear ratio.
Peak input power was 170 watts.
but required inertia for 1:1 was .12 oz*in*se^2
...4 times lower than our motor.
But inertia match does not have to be exact
... 0.25 to 4 is generally ok.
At .25 we have about 375 oz*in required.
Looks close!
Now 20:1
270 Peak input power
Still about 375 in* oz
But even at an inertia ratio of .25 we need
about 0.1 in*oz*sec^2. So there is no solution.
The treadmill's high inertia has killed it.
OK...done I think! With the derated motor we just
get by with a 10:1 gear reduction. anything much higher
or lower gets worse.
Max rpm is 382 so we can go much much faster than design
top speed if we want.
We are right at the comfortable motor current limit.
The motor will only need about 20V for design speed but
LOTS of current. This depends on the torque constant
but might be in the 10 amp range.
If the Pinion was one inch diameter rather than two,
5:1 would be optimum and so on.
Understand the results might be VERY different for
industrial servos and steppers...this is only for the
treadmill motors.
As a last bit...if any one has some of these motors
and a reasonable power supply I can show you how to
measure moment of inertia etc so we won't be doing
educated guesses.
I would like to know!
Les
Leslie M.Watts
L M Watts Furniture
Tiger Georgia
(706) 212-0242
Main page:
http://www.lmwatts.com
Engineering:
http://www.lmwatts.com/shop.html
Cnc surplus for sale:
http://www.lmwatts.com/forsale.html
Carved signs:
http://www.lmwatts.com/signwp.html
As promised I did an analysis of a hypothetical cnc router
with belt or rack drive using treadmill motors.
I had to do some eductated guesses...because there are few
specs on them. These are the cheap 1-2 hp 100vdc units
at surplus center and others.
First, I am going to seriously derate these things because they are
"treadmill" duty. I don't believe the HP ratings anyway. Let's derate
them 50%.
Second I have to get some idea of the rotor moment of inertia. None have
such specs, but I know a 3/4 hp square frame typical PMDC motor comes
out at about 0.4 oz*in*sec^2. I will use that figure. This is a very
high inertia compared to a typical industrial servo and will be one of the
biggest factors limiting performance.
Here is the router specs:
100 kG moving mass
125N (25 lb) cutting force
50 N (10 lb) friction force
100 mm/sec (4 inches/sec) max speed
.25g (2.45 M/sec^2) max acceleration
50mm (2 in) rack pinion/timing pulley diameter
Ok.
First the moment of inertia at the pulley/pinion.
That is simply Mr^2= 100*(.025M^2)= 8.8 oz*in*sec^2
A 1:1 load to motor inertia match occurs with SQRT( 8.8/.4)=
4.69 gear reduction ratio. Lets make it an even 5:1.
Starting with that I went to motioneering and
checked it with real motors.
I needed 600 in oz peak at only 171 rpm.
Peak input power was only 90 watts.
But....this is a no go because I have derated the motor.
I don't think they can put out much more than 400 in oz peak
and much less continuous.
Let's go with 10:1 gear ratio.
Peak input power was 170 watts.
but required inertia for 1:1 was .12 oz*in*se^2
...4 times lower than our motor.
But inertia match does not have to be exact
... 0.25 to 4 is generally ok.
At .25 we have about 375 oz*in required.
Looks close!
Now 20:1
270 Peak input power
Still about 375 in* oz
But even at an inertia ratio of .25 we need
about 0.1 in*oz*sec^2. So there is no solution.
The treadmill's high inertia has killed it.
OK...done I think! With the derated motor we just
get by with a 10:1 gear reduction. anything much higher
or lower gets worse.
Max rpm is 382 so we can go much much faster than design
top speed if we want.
We are right at the comfortable motor current limit.
The motor will only need about 20V for design speed but
LOTS of current. This depends on the torque constant
but might be in the 10 amp range.
If the Pinion was one inch diameter rather than two,
5:1 would be optimum and so on.
Understand the results might be VERY different for
industrial servos and steppers...this is only for the
treadmill motors.
As a last bit...if any one has some of these motors
and a reasonable power supply I can show you how to
measure moment of inertia etc so we won't be doing
educated guesses.
I would like to know!
Les
Leslie M.Watts
L M Watts Furniture
Tiger Georgia
(706) 212-0242
Main page:
http://www.lmwatts.com
Engineering:
http://www.lmwatts.com/shop.html
Cnc surplus for sale:
http://www.lmwatts.com/forsale.html
Carved signs:
http://www.lmwatts.com/signwp.html
Discussion Thread
turbulatordude
2005-05-16 09:09:48 UTC
where's the torque ?
Andy Wander
2005-05-16 09:46:50 UTC
RE: [CAD_CAM_EDM_DRO] where's the torque ?
R Rogers
2005-05-16 10:06:46 UTC
Re: [CAD_CAM_EDM_DRO] where's the torque ?
Leslie Watts
2005-05-16 11:16:11 UTC
RE: [CAD_CAM_EDM_DRO] where's the torque ?
caudlet
2005-05-16 14:04:35 UTC
Re: where's the torque ?
turbulatordude
2005-05-16 14:23:59 UTC
Re: where's the torque ?
Erie Patsellis
2005-05-16 16:34:45 UTC
Re: [CAD_CAM_EDM_DRO] where's the torque ?
Jon Elson
2005-05-16 18:20:47 UTC
Re: [CAD_CAM_EDM_DRO] where's the torque ?
Leslie Watts
2005-05-17 10:22:55 UTC
[CAD_CAM_EDM_DRO] treadmil motor speed reduction analysis
turbulatordude
2005-05-17 13:31:33 UTC
Re: treadmil motor speed reduction analysis
cnc_4_me
2005-05-17 15:34:00 UTC
Re: treadmil motor speed reduction analysis
R Rogers
2005-05-17 16:44:51 UTC
Re: [CAD_CAM_EDM_DRO] treadmil motor speed reduction analysis
Erie Patsellis
2005-05-17 17:06:16 UTC
Re: [CAD_CAM_EDM_DRO] treadmil motor speed reduction analysis
Leslie Watts
2005-05-17 17:15:44 UTC
RE: [CAD_CAM_EDM_DRO] treadmil motor speed reduction analysis
volitan712003
2005-05-17 17:31:30 UTC
Re: treadmil motor speed reduction analysis
Leslie Watts
2005-05-17 17:50:53 UTC
RE: [CAD_CAM_EDM_DRO] Re: treadmil motor speed reduction analysis
R Rogers
2005-05-17 20:24:57 UTC
Re: [CAD_CAM_EDM_DRO] treadmil motor speed reduction analysis
Erie Patsellis
2005-05-17 20:47:15 UTC
Re: [CAD_CAM_EDM_DRO] treadmil motor speed reduction analysis