Re: [CAD_CAM_EDM_DRO] motor speed/gear ration
Posted by
Jon Elson
on 2004-09-22 21:22:53 UTC
nielsenbe@... wrote:
selected
a motor already, find out what the peak torque is. Then figure out what
the worst
case cutting load you can imagine will be, and calculate back through
the leadscrew
to figure out how much torque you need to deliver that linear thrust.
The weight of the
machine parts and acceleration required are also important, especially
in router
applications. Once you
have the needed torque at the leadscrew, compare that to the motor's
peak torque.
This sets the minimum drive reduction. For instance, if you figured you
needed
250 Oz-In at the leadscrew, but only had a 125 Oz-In motor, then you
need a minimum
2:1 speed reduction.
depends TOTALLY
on the motor. For instance, the ubiquitous Ametek motors are rated for
900 RPM.
They can actually do more safely, but 2000 RPM might be pushing it.
weight gantry
just sitting on rails will not handle vicious accelerations well, as the
gantry may
lift off the rails. A well-constrained linear rail system that is
designed to eliminate
"racking" of the carriage, and has the thrust applied at the center of
mass can take
a lot of acceleration. Probably you should limit a homemade machine to
4 G's or so,
unless it is being built very solidly.
Jon
>I am building my first CNC system and I can't seem to find what speed servo'sThere is no "right" value. It depends on the motor. Assuming you have
>should be run at. I realize they can cover a wide RPM range but there must be
>an ideal range. I am trying to determine my physical design so I need to
>select a configuration/gear ratio that will work best for the motor.
>
>
selected
a motor already, find out what the peak torque is. Then figure out what
the worst
case cutting load you can imagine will be, and calculate back through
the leadscrew
to figure out how much torque you need to deliver that linear thrust.
The weight of the
machine parts and acceleration required are also important, especially
in router
applications. Once you
have the needed torque at the leadscrew, compare that to the motor's
peak torque.
This sets the minimum drive reduction. For instance, if you figured you
needed
250 Oz-In at the leadscrew, but only had a 125 Oz-In motor, then you
need a minimum
2:1 speed reduction.
>I will be moving a 35-50lb carriage on THK linear bearings, travel will beIn very general terms, the 2000 RPM will probably be better, but it
>~40" @ 250 inches per minute. I have considered using a .2 or .25 lead ball
>screw setup with a bayside reduction of ?? or possibly a rack/pinion setup due to
>the high IPM rate if I can find a zero backlash option.
>
>The pulling force should be rather low but I have no way of determining an
>exact number.
>
>I don't know if I should make the setup so it's running the servo at 200RPM
>or 2000RPM???
>
>
depends TOTALLY
on the motor. For instance, the ubiquitous Ametek motors are rated for
900 RPM.
They can actually do more safely, but 2000 RPM might be pushing it.
>Does anyone have an idea how much distance it will take to get the carriageIt again depends totally on the construction of the machine. A light
>0-250 ipm and then back to 0 on the other end without beating the system to
>death.
>
>
weight gantry
just sitting on rails will not handle vicious accelerations well, as the
gantry may
lift off the rails. A well-constrained linear rail system that is
designed to eliminate
"racking" of the carriage, and has the thrust applied at the center of
mass can take
a lot of acceleration. Probably you should limit a homemade machine to
4 G's or so,
unless it is being built very solidly.
Jon
Discussion Thread
nielsenbe@a...
2004-09-22 18:00:08 UTC
motor speed/gear ration
R Rogers
2004-09-22 18:54:36 UTC
Re: [CAD_CAM_EDM_DRO] motor speed/gear ration
Jon Elson
2004-09-22 21:22:53 UTC
Re: [CAD_CAM_EDM_DRO] motor speed/gear ration