motor size calculations

[drafterman]

I have a 36" x 48" router table that I am designing that is made of steel. I am trying to calculate the proper motor size I need. I will be using two motors to move the gantry along the long axis. I am thinking 425 oz-in steppers; coupled directly to a 1/2-10 ACME thread (1 start). I was hoping for around 60 ipm (motors may not turn that fast). What I did as far as a motion layout thus far is to say: if F=MA, then a 120 lb. (3.75 slugs) gantry with router travelling at 60 ipm (.08 ft/sec) will decelrate and accelerate in .1 sec (so 1.6 ft/sec/sec); thus, F=6 lbs. (seems very low to me...am I doing this right?)

Then, if F (oz)=2pi(motor oz-in)(screw eff %)/screw lead....using 35% to 40%
efficiency for my leadscrew, I get 9341 oz per motor or 583 lbs. per motor (seems way high...am I doing this right?)

Please advise as my high school physics memory is fading fast...

[mxtras]

You might find this helpful - look for the motion control download:

http://www.macphersoncontrol.com/navbar/mcptools.html

This is for a servo system but the results will be the same for steppers, etc.

Scott

[Zumba]

Although I didn't verify your calculations, they seem pretty accurate to me.

Keep this in mind.
1G = gravitational acceleration = 9.8m/s/s = ~20,000IPM/S = 2000IPM per 0.1 sec. You're doing like 1/35 of a G. Divide your gantry weight by 35 and that's how much force is required, ignoring friction.

ACME leadscrew efficiency is pretty abysmal, but at such low speeds, you're still only looking at a couple pounds of force. Twin 400-oz-in motors is totally overkill for 60ipm.

Just to give you an idea of what a stepper motor is capable of, I was able to take a 270oz-in NEMA23 motor @ 48v and drive a 150 pound gantry at 600IPM with much faster acceleration speeds. It had a ballscrew drive, but nonetheless...