What steppers do I need?

[MechanoMan]

Planning a gantry table router.
Looking at dual 5ft ballscrews (each with their own stepper and driver) with 16mm/turn motion. The table width is not nailed down right now, depends on whatever ballscrew shows up. 4 or 5 ft, ideally. It'd be framed with 8020 aluminum extrusion, including the gantry weight. Spindle weight unknown.

My question is, what stepper do I need? I was planning on driving without a reduction gear.

I found the equation for torque vs ballscrew linear force:
http://www.nookindustries.com/ball/BallGlossary.cfm
Td=P*L/2*PI *e
WHERE:
Td = Drive Torque (pound-inches)
P = Load force (lbs.)
L = Screw Lead (inches/turn)
e = Ball Bearing Screw Efficiency (90%)

Starting with a 495 oz-in NEMA23 motor (which seemed undersized), I get 277.5 lbf holding power for each side, or 555 lbs total. Is that right??

Now I know on the G540 these steppers perform well on the taig up to at LEAST 40 ipm, which is 800 rpm/2667pps on the steppers. That speed would be 500 ipm on this ballscrew. I was designing for 500ipm routing speed, so... good.

Glancing at Keling's stepper charts (which probably don't use as good a drive as the G540), the torque drops off to about 77% of holding torque at 2666ppm. But that's still 432 lbs of force at top speed. Even if the gantry and spindle weighed 20 lbs, it'd pull 21.6 G's before stalling.

The Y axis would have only one stepper. But, considerably less inertia, given it's only moving the spindle and Z-axis, and not the entire gantry.

I'd like to reuse the G540, smoothstepper, computer, and power supply off my existing mill, and just plug in the router steppers when I need the router. I won't be running both at once. This would save me a lot of money. The G540 cannot drive the higher currents of the NEMA34 or NEMA42. Like I say, the NEMA23 SEEMED too small for the task- but the calculations sure don't seem to bear that out.

Huh, I did not consider the rotational inertia of the ballscrew shaft itself though when estimating acceleration.

Is this sound? Can a pair of 495 oz-in NEMA23's push a large router gantry at high speed?

[ger21]

Couple things. Nook tells you to use the "torque to move 1 lb" # in their catalog, as it's easier than the formula. They don't seem to have a 16mm lead screw, so I looked at inch screws with 1/2" lead and 1" lead. The 1/2" lead needed 1.4 oz-in to move 1 lb, the 1" lead needed double that (2.8). That means it's pretty linear. 16mm is roughly .625", so, 1.4/4*5 = about 1.75 oz-in of torque to move 1lb. So using that, I get 495*1.75 = 282 lbs of force, which is about the same as you got.

Problem lies with your Keling data. I didn't see a torque chart for the 495. You can assume all charts are the same. It appears that the charts are using 1/2 step, so the 2666 pps = 400 rpm, not 800. Also consider that different drives will give different curves. And last, the 495 oz motors are not known for high rpm power. I'd be surprised if it had 30% of it's holding torque at 800 rpm.

Even if the gantry and spindle weighed 20 lbs, it'd pull 21.6 G's before stalling.

I wouldn't be surprised if your gantry weighs 100-150 lbs. My mdf and plywood gantry weighs about 78 lbs with the spindle.

But, say you do have 400+ lbs of force, and a 20lb gantry. If you accelerate at 21G's, as soon as the force drops below the 400+ lbs needed for that accel rate, it'll stall. Which would be almost instantly. What you need to look at, is how much force you have at your top speed. That's the force that you have to use to calculate the acceleration. Acceleration is constant, so you need that same force available until you reach your top speed. Screw diameter can play a huge role in accel as well.

And I think if you can even accelerate at 1-2g, that's pretty fast. My routers accel is set to .025 G's, and even that is pretty quick up to the 150ipm max.

Just off the top of my head, doesn't 21 G's get you to about 8000ipm in 1 second? I just checked, and yes, it is.

What you really need to do, is figure out how much force you need to have at 500ipm, and how much force you need to accelerate at a given rate, then choose a motor that fits.

[MechanoMan]

I did use the chart for the NEMA34 stepper KL34H260-60-4B listed above the NEMA23 I needed a chart for, and just scaled it. It does say "half step". The G540 may well be a better drive and show less loss at higher speed than the chart indicates.

OK, thanks for confirming my calcs- I was thinking that surely I was applying the formula wrong, but you've got basically the same answer coming from another method.

OK, the "pulses per second"... I see, I totally misread that. That would bring the 500ipm torque down to ~50% for the NEMA34 chart (which isn't the motor I'm looking at).

OK, I may well have seriously underestimated the gantry weight.

I believe the acceleration was correct. If the gantry weighed 20 lbs and Mach3 was configured for 20G's, then (neglecting rail friction losses and milling force) it would need 400 lbs of force at every speed it goes through or it would face a stall at that speed.

Of course 21G is plain absurd and would not actually be used or needed. The frame would leap across the floor underneath it! No the only reason I mention it is the calculation suggested "there is many times more force available than I would ever need", which is always good. If the Keling NEMA23 495oz-in DID actually lose a great deal more torque at 500ipm than I got from the other chart, it may still be enough.

So from the NEMA32's chart, that would STILL have 277.5lbs of linear force. If it were a 100lb gantry still accelerating at 0.25G as it got to 500ipm, I'd still have 253 lbs of force available for cutting and overcoming rail friction. How much IS cutting force, like 1/2" carbide at 10K rpm into maple at 0.25" depth? I don't have a frame of reference here. 0.25G would take it from 0 to 500ipm in 0.02 sec so I think we're already more than I'd need.

BUT, again, that's not exactly the stepper I'd be using. OK, Keling DOES have a chart for the 425oz-in one. And it's a quite different story at 500ipm speed- more like 20% of its holding torque.

Oh wait- there's a chart for the NEMA23 387 oz-in. And (after noting this is 1/8 steps so I need a 2000pps point for 500ipm) THAT one shows about the same 0.9N-M torque at 500ipm rate I initially calculated for from the NEMA32 chart.

I wish they had charts based on use with a 48V G540 drive!
What sort of motors do people normally use on a ballscrew router of this size? Do they ever use NEMA23 steppers?

[ger21]

How much IS cutting force, like 1/2" carbide at 10K rpm into maple at 0.25" depth?

At 100ipm or 500ipm? If you're only cutting 1/4" deep, not much. It does depend on the type of bit as well as the sharpness of it.

I like to use 30-50 lbs for cutting force when doing calculations. You shouldn't see more than that with a 3HP or smaller spindle.


And yes, the 21G's was correct. I didn't believe it at first, but did the math.

[MechanoMan]

Lemmie make sure again, with the 387 oz-in chart:
500in/min *(1rot/0.63in)*(200steps*/rot)*(1min/60sec)=2646 steps/sec
with 1/8 stepping= 21164pps

Chart for the actual NEMA23 387on-in:
http://www.kelinginc.net/KL23H284-35-4BT.pdf
=0.9N-m @ 21Kpps.

0.9N-m*(141.6oz-in/N-m)=127oz-in Wait, this is MUCH LOWER than I expected- somehow the 0PPS value on that chart only translates into 297.4 oz-in, not the 387oz-in rating! This is a major difference!

127oz-in=7.9375lb-in

Td=P*L/2*PI *e

P=Td*2*3.14*0.9/0.63
=71 lbs per side

142 lbf total from ballscrew drive @ 500ipm

100lb gantry at 0.1G (seems likely):
500ipm*(1ft/12in)*(60sec/min)=0.69ft/sec
0.1G*( (32ft/sec*sec) / 1G)=0.217sec to go 0 to 500ipm, sounds "fast enough" to me.

100lb*0.1G=10lbf needed to overcome inertia
50lbf cutting resistance
142/(50lbf+10lbf)=2.36x more torque than needed

142-(50-10lbf)=82lbs surplus torque

The Keling chart was "probably" not for as good a drive as the G540.

So the second version of the calcs is not AS good, it's not "so insanely higher than needed you can stop worrying ever again"- but still shows there's a GOOD safety margin against stalling here, right?

I have concerns about backdriving. This chart is for loading with torque force opposite the direction it's stepping in, moving against a resistance. But, in a ballscrew, attempting to slow down suddenly creates a torque force on the stepper shaft in the same direction as it's trying to turn. I don't think it's outright safe to assume the torque produced in this arrangement is the same as the load Keling characterized in its doc.

I guess the next step is to contact Geckodrive about how this backdriving "works" on their drives.

[MechanoMan]

I was interested to note that based on that stepper chart:
http://www.kelinginc.net/KL23H284-35-4BT.pdf

Changes in thread pitch don't change the resulting linear force at various speeds!
For example, even if I were to find a screw with HALF the pitch, that would require half the torque, but twice the rpm shaft speed. But looking at the chart, the torque at 40KPS is pretty much EXACTLY half the torque at 20KPS. And that's the same story if I were to add a 2:1 reduction drive on a screw.

What does change somewhat though is the resolution. At 16mm pitch, each fullstep in 3mm. Microsteps are not entirely linear. So there's "kinda" a possible accuracy issue there, but probably not.