[Bear5k]

Point is, you can't just slap another higher torque motor in, and expect better rapids (though you might have better performance for cutting speeds.)

I'm not sure how you get from 1N-m to 700oz-in, but you're right. In all of that, I was just advocating slapping any old random motor with a higher torque rating into the system.

Evevrything else must be taken ito cosideration.

As I said, I don't have anything more to contribute. Best of luck to all.

[louieatienza]

I'm not sure how you get from 1N-m to 700oz-in, but you're right. In all of that, I was just advocating slapping any old random motor with a higher torque rating into the system.


As I said, I don't have anything more to contribute. Best of luck to all.

1 N*m is about 140 in*oz, 5mm ballscrew. Not hard to figure out. I think Archimedes figured this out 4000 years ago or somewhere....

While the information you provided is very informative, I believe we all get a little carried away at times wit too much information, when a little bit of common sense and maybe some simple algebra, mixed with some practical knowledge, can be more helpful than a barrage of facts and data, backed by some self-agrandizing attitude. We're all trying to help here, and I'm sure all information is appreciated.

[Bear5k]

1 N*m is about 140 in*oz, 5mm ballscrew. Not hard to figure out. I think Archimedes figured this out 4000 years ago or somewhere....
[..]backed by some self-agrandizing attitude[..]

Archimedes, huh? Let's run what my simple, tired, newbie brain can figure out, and then I'll hand it over to you:
500ipm = 211.67mm/s
211.67mm/s = 42.33rps (@5mm/rev)
42.33rps = 8,466.64 pps (@200pulse/rev)
(quick look-up to the charts above)
The big motor is at ~0.8N-m and the smaller motor is at ~1N-m.

So, now we need to get from the above scenario where each motor is notionally putting out no more than 140oz-in to where they are cranking out 700oz-in. It is just simple 4,000 year old algebra, so teach away!

Remember, this was your starting point (emphasis mine):

So let's say you have a driver/computer capable of spinning the Sanyo Denki motors at full potential. With the 5mm leadscrews, just off the top of my head, you'd be pushing the gantry with about 700 in*oz of force, at about 500ipm, if your ballscrews can handle it.

As I said, best of luck to all! :cheers:

[louieatienza]

Archimedes, huh? Let's run what my simple, tired, newbie brain can figure out, and then I'll hand it over to you:
500ipm = 211.67mm/s
211.67mm/s = 42.33rps (@5mm/rev)
42.33rps = 8,466.64 pps (@200pulse/rev)
(quick look-up to the charts above)
The big motor is at ~0.8N-m and the smaller motor is at ~1N-m.

So, now we need to get from the above scenario where each motor is notionally putting out no more than 140oz-in to where they are cranking out 700oz-in. It is just simple 4,000 year old algebra, so teach away!

Remember, this was your starting point (emphasis mine):


As I said, best of luck to all! :cheers:

The lead of the screw acts like gear reduction, so the torque of the motor is multiplied by the TPI of the screw, and the speed divided by the TPI of the screw. This of course does not take into account any friction or other inefficiencies in the system, but it gives a place to start.

This is why I believe barefootby4's machine may be stalling due to resonance. Even if his x axis stepper only had a third of the it's holding torque at 500rpm, that's still almost 1000 in*oz of force at 100 ipm, which should overcome any slight binding, or even an uncooperative ballnut. I don't know if you could run a typical stepper over 500 rpm reliably at it's rated volatage, unless you use a higher powered power supply, which was also suggested.

Now, using your example above: unless you're using that Sanyo Denki stepper (which has a rated max rpm of 3000), or a servo, you're not going to spin that Chinese ballscrew at over 2500 rpm, even with two fixed end bearing blocks. I have some precision ground C5 ballscrews that have a lower max rpm than 2500, and the Chinese ballscrews are C7 rolled, with C7 (supposedly) bearings. This is not a knock on the Chinese bearings/blocks, as there is a place such products at their price/performance/quality point. There in use in many machines and their owners are happy. I own a set of three "C7" fixed/free end bearing blocks purchased from the same guy, and it only takes a quick glance at them in hand to determine that spinning those bearings at 2500 rpm might nit be a good idea. The bearing on the free end is marked "NSK" but the machining on the fixed end block is sloppy, and the retaining nut seems to rub on the bearing plate.

Using your own data, you've determined that the smaller motor has better performance at about 2500 rpm, than the larger one. I'm not saying there isn't a place for the larger stepper. For a large r&p system with a heavy gantry, and high HP spindle for cutting sheet stock at high speed in one pass, it's needed. I believe for the application here, it might be way overkill.

As for your algebra you could have made it simpler:

Divide the feedrate by the screw lead to get the screw rpm:

500ipm / (5mm /25.4[mm/in]) = 2540rpm (approx)

Then you can divide by 60 to get the revolutions per second:

2540rev/m / 60 sec/min = 42.33 rev/sec

Conversely you could just convert the ipm to ips:

(500inch/min / 60 sec/min) / (5mm /25.4[mm/in]) = 42.33 rev/sec (guess my algebra is not to shabby)

So as long as your computer can handle over 500,000 ppm, AND you have a Sanyo Denki Step-Syn Stepper, AND you ballscrew can handle over 2500 rpm, THEN maybe you can see 500 ipm [with a 5mm lead ballscrew]. Even so, I have a feeling you wouldn't be able to set you accel/decel high enough to reap the benefits.

[Bear5k]

Please refer to page A-718.
http://www.thk.com/documents/us_pdf/...11studying.pdf

The torque required to turn a ball screw is simply the load multiplied by the angular velocity. You see "gearing" in a multi-start screw because the effective lead is different than the stated lead. With a ball screw, you don't have this "advantage" (really an obfuscation), and any gearing would need to be just that -- some form of m:n reduction gearing/pulley system. If I am wrong, I would love to see some references, since I'm missing something obvious or missed a class too many back in undergrad days, and I need to get deeper background information (lead or TPI is factored in via the conversion from linear velocity to rotary velocity).

We agree on the question of whether 500ipm is feasible with a 5mm lead on a Chinese screw. Critical speed on my C5 NSK screw is around 2500rpm (I'd have to look it up), and this will change with load (resonance). I am looking to replace it with a longer screw, also from NSK, and that one's critical speed is a bit south of 2300rpm. Both are 10mm lead, so I'd be doing 500ipm at about half the critical speed of the unloaded screw. For the OP, 500ipm is probably at or above his critical speed taking load into account. Let's try to get him to 200ipm or so.

I also agree that resonance is probably the issue, though there would be others that I would check as well just to rule them out. If it is resonance, the OP will either need to fabricate a mechanical damper or invest in a drive unit that can do this electronically (e.g., Gecko, Parker, etc.). He could also try to "power through" the resonance (skip up to an even higher speed that is outside of the resonance band), but I do not know how successful this strategy really would be. Changing the load on the motor (mostly by changing the cutting load since the gantry load would be fixed) might change the resonance enough, as well, to power through it. Each of these latter options are relatively easily tested and cost-free except for the time and effort.

If it isn't a resonance issue (or one of the other mechanical issues), but an issue with the load itself (e.g., back to "feeds and speeds") that the OP is using or needs to use, then the checkbook is going to need to come out for some combination of new motor(s), new drives, new breakout boards or new power supplies. He could try to increase the voltage (really current) going to the motors since he is running 50V when most of Keling's motors seem to be rated at 100V, but I suspect the drives he is using won't handle the increased current (the 'big' Gecko's only go to 80V).

If the OP is having a resonance issue that can't be solved with a mechanical damper, then he could be chasing a troubleshooting process that leads to swapping out a lot of electrical gear without a lot of incremental performance improvement versus what he could have gotten through a more targeted approach (death by 1,000 cuts). This would be both expensive and tragic (I've got my own history of correcting design oversights/mistakes and will most likely make many more, myself). He should figure out how much his budget is, how much the current problem bothers him, and then look at a more complete solution. If "complete solutions" are inside his budget, then there is a logic to just going to the next performance level, saving the incremental troubleshooting/debugging time and expense. However, if he is budget constrained, there are several good and cheap options on the table to try first.

One motor option that would be a bigger motor with more torque across the board (sticking with Keling) is this one:
http://kelinginc.net/KL34H295-43-8BT.JPG

It is also rated at 100V, so performance would suffer at 50V. As a result, one would only really want to swap into that one if one is also changing out the drives, the PSU and probably the BOB, too. Hopefully the OP has enough information now to go back and check the suitability of his gear to his intended real-world usage pattern and see where the yellow and red flags are raised. Troubleshooting can then guide him to where to focus and on which parts.

[louieatienza]

Please refer to page A-718.

The torque required to turn a ball screw is simply the load multiplied by the angular velocity. You see "gearing" in a multi-start screw because the effective lead is different than the stated lead. With a ball screw, you don't have this "advantage" (really an obfuscation), and any gearing would need to be just that -- some form of m:n reduction gearing/pulley system. If I am wrong, I would love to see some references, since I'm missing something obvious or missed a class too many back in undergrad days, and I need to get deeper background information (lead or TPI is factored in via the conversion from linear velocity to rotary velocity).

There are ACME screws with different leads and numbers of starts, but the same effective TPI. For example, there is a 1/2"-8, 4 start, and a 1/2"-10, 5 start. Though the "leads" may be different, they both have the same effective TPI, and my guess is that they'll preform almost the same, save for the inefficiency of the AB nuts. In fact, they would perform as if there were only two threads per inch, and an AB nut could be made with the same coefficient of friction as the above.

Now you could get a ballscrew with say a 1/2" lead, and get the same performance as the ACME screw with a smaller motor simply for the fact the ballscrew/nut would be way more efficient than tha ACME screw.

I wasn't able to download the pdf (link broken?), do you have the specific formula to post here?

PS Let's leave the obfuscations to the piliticians!

[Bear5k]

Try this one instead. I remembered to save a copy of it to my hard drive in case this one is broken within hours of my referencing it.

https://tech.thk.com/en/products/pdf/en_a15_057.pdf

[louieatienza]

Try this one instead. I remembered to save a copy of it to my hard drive in case this one is broken within hours of my referencing it.

https://tech.thk.com/en/products/pdf/en_a15_057.pdf

Got it...

I still think however, using simpler math, that it doesn't matter whether it is a ballscrew or an ACME screw. Let's pretend the ACME screw has the same efficiency as the ballscrew. If they have the same effective tpi, let's say 5, and using the same motor, both motors would do 5 times the amount of work over a given distance than a screw with a 1:1 tpi/lead ratio. Of course, the ballscrew could do the same amount of work as the ACME screw, with a smaller motor, due to its higher efficiency.

So I believe, disregarding friction, that the screw lead multiplies (or divides) the availble torque of the motor just as if you had geared down the motor to a screw qith a 1:1 tpi/lead ratio.

Speaking of leads in ballscrews, there are "freewheeling" ballscrews, whose effective lead changes depending on the load at the nut. Also, roller screws can have an effective lead higher or lower than the screw lead, just by changing the roller nut. In fact, a roller screw can have a nut with opposite-handed threads in the rollers, or rollers larger or smaller than the screw diameter, which can affect the effective lead, without changing the screw!

[Bear5k]

Smaller lead screws are less efficient (more binding) than moderate lead screws, but as I indicated above, my understanding is that once you convert the linear (axial) motion into radial motion via the lead, its relevance disappears. If a motor supplies 1 N-m of torque at a given angular velocity, then that is how much torque you have. To increase torque, you can gear the motor, but then the motor is operating at a higher angular velocity than the screw.

I only bring up the special circumstances of ACME screws because of multi-start screws and how people interpret (or misinterpret) them. Once you get a multi-start screw converted to its actual TPI, then there's no gearing (but a lot of drag/friction vs. a ball screw).

[louieatienza]

Smaller lead screws are less efficient (more binding) than moderate lead screws, but as I indicated above, my understanding is that once you convert the linear (axial) motion into radial motion via the lead, its relevance disappears. If a motor supplies 1 N-m of torque at a given angular velocity, then that is how much torque you have. To increase torque, you can gear the motor, but then the motor is operating at a higher angular velocity than the screw.

I only bring up the special circumstances of ACME screws because of multi-start screws and how people interpret (or misinterpret) them. Once you get a multi-start screw converted to its actual TPI, then there's no gearing (but a lot of drag/friction vs. a ball screw).

OK I'll give another example. Let's say you're using a servomotor, which can supply constant torque thoughout it's rpm range, say 0 to 3000 rpm. Now you have two ballscrews, one with a 5mm pitch and a 25mm pitch. You spin each screw so that you're producing the same feedrate (5mm screw spins 5x as much.) You're saying that both systems do the same amount of work per unit measurement?

[Bear5k]

OK I'll give another example. Let's say you're using a servomotor, which can supply constant torque thoughout it's rpm range, say 0 to 3000 rpm. Now you have two ballscrews, one with a 5mm pitch and a 25mm pitch. You spin each screw so that you're producing the same feedrate (5mm screw spins 5x as much.) You're saying that both systems do the same amount of work per unit measurement?

Now you are getting into a more complicated question (system-level vs. motor). Assuming the same payload for each and decent quality servo drives, the motor attached to the ball screw with the 5mm lead will be producing (and consuming) more power than the one with the 25mm lead -- most likely -- but this will be almost entirely due to the increased friction in the 5mm lead and the accumulated electrical inefficiency in the motor spinning at higher RPM. This last bit is a guess based upon the motors I've researched. Since it would be a result of practical implementation, it's a testable hypothesis rather than a derivable fact.

In pure theoretical physics, they would be doing the same amount of work (force times distance) and consuming the same amount of power (work over time; force times distance divided by time). Think about it this way:

To get to the same linear speed, the 5mm lead screw has to turn 5x as much, but with only 1/5th the load per revolution. The 25mm lead screw would need to turn 1/5th as much, but it will need 5x the torque per revolution. This may be the source of the gearing question?

The good thing about servos is that they will only consume as much power as needed, up to their limits (they are more efficient than steppers, which pump out a lot of waste heat). Thus, if you sized the motors for the torque requirement of the 25mm lead and the angular velocity requirement of the 5mm lead, you would see both screws operating well within their limits. The 5mm lead being able to take on more payload (5x more) at that same speed, while the 25mm screw could take on more speed (5x as much) at the same payload.

Make sense?

[ger21]

Smaller lead screws are less efficient (more binding) than moderate lead screws............

If you download Nook's acme catalog, they list the efficiency of all their screws.

If you compare screws with 1 turn/inch, screws from 1/4" to 1" diameter are all between 75%-80% efficiency. Larger than that, and the efficiency drops quite a bit.

Maybe I don't understand what you're saying?

[louieatienza]

Now you are getting into a more complicated question (system-level vs. motor). Assuming the same payload for each and decent quality servo drives, the motor attached to the ball screw with the 5mm lead will be producing (and consuming) more power than the one with the 25mm lead -- most likely -- but this will be almost entirely due to the increased friction in the 5mm lead and the accumulated electrical inefficiency in the motor spinning at higher RPM. This last bit is a guess based upon the motors I've researched. Since it would be a result of practical implementation, it's a testable hypothesis rather than a derivable fact.

In pure theoretical physics, they would be doing the same amount of work (force times distance) and consuming the same amount of power (work over time; force times distance divided by time). Think about it this way:

To get to the same linear speed, the 5mm lead screw has to turn 5x as much, but with only 1/5th the load per revolution. The 25mm lead screw would need to turn 1/5th as much, but it will need 5x the torque per revolution. This may be the source of the gearing question?

The good thing about servos is that they will only consume as much power as needed, up to their limits (they are more efficient than steppers, which pump out a lot of waste heat). Thus, if you sized the motors for the torque requirement of the 25mm lead and the angular velocity requirement of the 5mm lead, you would see both screws operating well within their limits. The 5mm lead being able to take on more payload (5x more) at that same speed, while the 25mm screw could take on more speed (5x as much) at the same payload.

Make sense?

This was exactly the point I was attempting to make, though maybe your choice of words might have been better than mine. Disregarding friction, you could use a motor with 1/5x the power to spin a 5mm pitch screw, to move the same amount of "payload" as with a 25mm pitch screw, at 1x power.

I disagree however that they'd be doing the same amount of work. Because the work done would be the same as if you took the threads from each screw, and unwound them as if it were a line. Thus the higher pitch screw would have a shorter "line" than the lower pitch screw; or it's doing less "work" over it's "radial" distance [not speaking in scientific nomenclature, of course]. But since we know that a stepper is being used, we know that it's torque drops as the rpm increases. But since the torque curve of the larger stepper drops more sharply than the smaller stepper at the same speed, the larger stepper may be less "stable" at that speed. So IMO it would be a waste to get the larger stepper, as the performance in rapids would be similar, unless you needed more low-speed power, which we already have in abundance. Also to consider is the cost of a ballscrew swap versus a motor swap. I think you'll get more performance for the money out of a screw swap versus a motor swap. And the old screw is probably an easier sell on eBay. I've crashed 1/4" bits and snapped them clean off with my 425 in*oz stepper, with 1tpi effective lead. Our guy here has over 5x the "pushing power" of my machine!

[ger21]

I disagree however that they'd be doing the same amount of work. Because the work done would be the same as if you took the threads from each screw, and unwound them as if it were a line. Thus the higher pitch screw would have a shorter "line" than the lower pitch screw; or it's doing less "work" over it's "radial" distance

If it's using 5x more force, than it's not doing less work, is it?

Which is less, carrying 10 pounds for 5 miles, or 50 lbs for 1 mile??

. But since we know that a stepper is being used, we know that it's torque drops as the rpm increases....................................... Our guy here has over 5x the "pushing power" of my machine!

Only when it's not spinning, or slow speeds. At higher rpm's, he most likely has less. Here's a good explanation of it.
Mechanical Power - PMinMO.com

[louieatienza]

If it's using 5x more force, than it's not doing less work, is it?

Which is less, carrying 10 pounds for 5 miles, or 50 lbs for 1 mile??



Only when it's not spinning, or slow speeds. At higher rpm's, he most likely has less. Here's a good explanation of it.
Mechanical Power - PMinMO.com

I think the question is, which is less - carrying 10 pounds to a point 5 miles away, going in a straight line, or carrying a 10 pound load to a point 5 miles away, but zig-zagging 25 miles to get there?

As to the 2nd point, that's exactly what I'm trying to illustrate. Even though his stepper is larger than mine, he'll have more power at slow speed, but his rapids performance will be less. That's why I don't believe a larger stepper is the solution.

[ger21]

I think the question is, which is less - carrying 10 pounds to a point 5 miles away, going in a straight line, or carrying a 10 pound load to a point 5 miles away, but zig-zagging 25 miles to get there?

The issue is that the load changes with the screw pitch.

[Bear5k]

If you download Nook's acme catalog, they list the efficiency of all their screws.

If you compare screws with 1 turn/inch, screws from 1/4" to 1" diameter are all between 75%-80% efficiency. Larger than that, and the efficiency drops quite a bit.

Maybe I don't understand what you're saying?

I'd have to dig up the references. I haven't played with Nook's stuff in a long while (I have THK and NSK screws, so I mostly troll through their tech bulletins these days), and I never paid much attention to the efficiencies of lead screws (ACME/Trapezoidal) with the general design guidelines I've seen quoting 50% or less (e.g., Kollmorgen, Parker).

For ball screws, the differences are minor, but potentially material to someone engineering to tight parameters (hopefully not a DIYer). For a ball screw the efficiency differences are driven by the angles needed to recirculate the balls. Small leads require more "aggressive" angles which are lower efficiency, but then the really long leads start to present their own issues that I've never really delved into, but seem to exist. I never graphed it, but I suspect that there is a sweet spot in the 10 - 20mm range for lead that is "optimal" for a ball screw. That being said, when you are dealing with 90%+ efficiency, we are talking about minor differences -- enough that implementation-specific issues will be more important.

Frankly, they could have the same efficiency for all that it matters. The meat of the point is that the work and power are the same across the two examples, which we seem to agree.

As to the 2nd point, that's exactly what I'm trying to illustrate. Even though his stepper is larger than mine, he'll have more power at slow speed, but his rapids performance will be less. That's why I don't believe a larger stepper is the solution.

Guys - I'll concede the point. If you aren't willing to give a bigger motor more current to do its job and to overcome the limitations of higher inductance, then a bigger stepper motor isn't likely to be a good answer under almost any circumstances. There are examples of decently-spec'ed motors out there, but when holding current constant, lower inductance (and thus, lower holding torque) is the better option. This is especially true if you are running a motor at a fraction of its design power (e.g., 100VDC motors run at 50VDC). There are some "gems" out there, but there is a lot more product on the market that is built to a cheapest level that can be squeezed into a spec if you can look at it cross-eyed long enough.

[ger21]

There are examples of decently-spec'ed motors out there, but when holding current constant, lower inductance (and thus, lower holding torque)

Lower inductance does not necessarily go hand in hand with lower holding torque.
But to get high torque, low inductance, you need to get into the 6A range.

[louieatienza]

The issue is that the load changes with the screw pitch.

This I think is where I'm getting a bit confused.If the load [the gantry] remains the same, then that doesn't change. The only that changes is the amount of force applied to it over a given distance.

So if the torque applied to the screw is T=Fs (where s is the circular length the screw turns) then [given the same motor running at the same rpm for both screws] the 5mm pitch screw would have 5 times the torque as the 25mm pitch screw, to move the gantry the same linear distance, since its circular distance is 5 times that of the 25mm screw.

[ger21]

This I think is where I'm getting a bit confused.If the load [the gantry] remains the same, then that doesn't change. The only that changes is the amount of force applied to it over a given distance.

Correct. More force is required with the higher lead screw, so more work is being done.

I think if you had a machine with servo motors, and an ammeter, you'd see the motor draw more current with higher lead screws, for a given load, and all else being equal.