Motor size in relation to workload

[woodworker2]

Hi, I have a small cnc router that I am currently using, (mostly for drilling purposes), and I am considering building a larger 5'x9' machine to cut plywood and expensive veneered sheetgoods on, mostly on 3/4" material and long 8' cuts. My question is this:for example a nema 23 stepper motor with 269oz./in. What does that mean? What is the relationship between motor size and the workload? Is there a formula to figure out motor size to the load being cut by a cnc? Thank you in advance for your responses.

[pminmo]

Typically the rating is holding torque. Torque being Force x Moment Arm. http://en.wikipedia.org/wiki/Torque But in all likelyhood you will convert the rotation of a motor to a linear means via a screw or gearing. That conversion will have an efficiency, as well as a mechanical amplification or attenuation of power depending on the ratio. That mechanical conversion is as important as the motor power.

[sansbury]

To cut a material properly you need to be able to push the cutting tool against it with sufficient force. There are ways to calculate this, but most people operate from rules of thumb since the number of screw/motor combinations is limited. The bigger limit you're likely to hit is rapid speed. The faster the motor turns, the less torque it produces. At some point you stall. A 269oz motor can provide plenty of force on 10/20TPI screws but it won't give you 200ipm jogs. I would look at commercial machines similar to what you want. In that size range I think you'll mostly see rack-and-pinion drives with 450oz+ motors. 269 might work but it might not move too fast.

[Oldmanandhistoy]

Have a read through this thread especially posts by Mariss.

http://www.cnczone.com/forums/showthread.php?t=31815

John

[energyforce]

You could put the a bit you want to use on your router and put it on the router on the wood you want to cut and then turn it on. Push the router at the speed you want it to cut. How much force does it feel like you are pushing sideways? You could try using a spring scale (you hold the router and your friend pull the spring scale).

This is the force you will need to push the router ONLY! Call this the router force. Heck multiply it by 2 or 3 because you want something strong.

As for the rest of the machine (gantry weights, friction, inertia from the leadscrews, acceleration etc) thats a different story/calculation. Once you find the force required for all that stuff, add your router force to this force.

Then change this force into the torgue you will need and see if 269ozin motor can push it. I think you will definately be okay with 269 ozin if the table is engineered properly. My table is friggin huge and friggin heavy and friggin fast and im using a 400 or 500 oz in motor.

Erik

[Al_The_Man]

Break-away torque, Movable weight and side cutting forces are only a part of the equation, you will probabally find that the greatest demands come with inertia required for acceleration/deceleration rates.
There are plenty of sizing programs out there that reduce the need for 'rule of thumb' to a minimum.
Al.