Basic servo motor size calculations (for aluminium)

[ftec]

Hi,

(This question has probably been asked and answered in some other thread but I just can't find it).

Some ten years ago I was involved in robotics design and we used to have a program to calculate the size of servo motors needed. This was a general program for assembly robotics with constant load on the arm or axis. Does anyone know of such a program (or even manual calculations sheets?) which would help to determine the size of servo motors needed for milling, aluminium, wood etc.

The reason I need this for is that I'm building a CNC machine for myself with the intention to use it also for 5 mm aluminium milling except for the normal woodworking purpose. For this reason the gantry is becoming fairly heavy, over 45 kg and I am worried that I am aiming at too small servo motors or too slow speed for wood working. I'm able to calculate manually the inertias etc. but I have no idea what are the forces needed to cut aluminium and the acceleration/response times need to keep an axis on the "milling cut track" (dunno the correct term for this in English) ie. to follow accurately the cut path, to make accurate cuts, if you know what I mean. Anyone able to help?

Thanks,
Risto

[irving2008]

Hi,

I am working up a spreadsheet (might even end up a C# program) to do just this, but some good sources of info are:
http://www.hmkdirect.com/downloads/G...ive_sizing.pdf

http://www.roton.com/application_engineering.aspx

http://www.roton.com/formulas.aspx

http://www.etec.wwu.edu/faculty/McKe...m%20charts.pdf

http://medmaninc.com/rbs/pdf/rbs-cat...rew-sizing.pdf

http://www.sizingtools.com/

I haven't found a specific answer to how to do load calcs for miling so I was trying to work it out from first principles but don't have an answer as yet... this site comes close: http://www.mapal.us/calculators/mill...torMilling.htm

Hope these help some...

Irving...

[ftec]

Thanks Irving,

You would be doing a great service for many I believe. One can save a lot of money and time with less hassle testing motors and drives etc.

I also found one free program and a book of the subject but an editable spreadsheet poking around here in the group would probably be very good for folks with different systems and purposes. Here's the program I found, haven't tested it yet:

http://www.copperhillmedia.com/Visua...gArticles.html


Have to go to work now, will check your links later.

Regards,
Risto

[Al_The_Man]

Most of the servo/stepper manuf. offer a graphic motor sizing program, one of the best was kollmorgen, but since being taken over by Parker, I am not sure it is still available, I put a link in the utilities section for an older Emerson one.
http://www.divshare.com/download/4412628-dac
They almost all base their sizing around the industry custom of sizing the motor to load inertia as being max. 10:1 or less, the biggest influence on this is your accel/decel rate for a given load.
I posted a PDF here some time ago of an article on motor-load inertia.
Al.

[irving2008]

Before I start let me say I am no expert, but I can read and interpret research papers and the like... however I may be wrong in my interpretation.

The determination of cutting force is non-trivial, having now read several research papers, none of which seem to agree to any great degree!

The cutting force is a non-linear function of axial and radial cutting depth (assuming an end mill), the feed rate (in/min or mm/min) and the cutting rate (in/tooth or mm/tooth) which is itself a function of revs, feed rate and number of flutes. Another factor is that for light cuts only one tooth is on contact while deeper (radial) cuts put more teeth in contact. The axial and radial depths govern the cut volume (material rate of removal which is analogous to energy used) and the surface area of contact which relates to friction and temperature as a function of feed rate and revs.

There is a lot of experimental data which results in a number of equations. The problem is that mostly these equations only work for the conditions of the experiment - if you plug in values outside that then they go awry. FOr example one equation works great for a 10mm cutter between radial and axial depths of 0.2 - 0.6mm and feed rates 50-150mm/min but fails completely when compared to other data for cuts of 1mm radial, 10mm axial @ 125mm/min

Some of the experimental data gives these numbers from which one might mentally extrapolate (these are 2 examples of about 5 or 6):


RD = Radial depth mm
AD = Axial depth mm
VC = Cutting speed mm/min
RPM = Spindle speed

Material: Steel ck45, 10mm 4-flute
RD AD VC RPM Force (max N)
0.2 0.2 125 250 115
0.2 0.2 125 275 108
0.2 0.2 125 300 102
0.2 0.2 125 325 97
0.2 0.2 125 350 93
0.2 0.2 125 375 88
0.2 0.2 125 400 83
0.6 0.6 250 500 353
0.6 0.6 250 550 329
0.6 0.6 250 600 315
0.6 0.6 250 650 304
0.6 0.6 250 700 295
0.6 0.6 250 750 287
0.6 0.6 250 800 279


Experiment 2 (different researcher)

Material: Steel AISI 1045, 10mm 4-flute
Force given as maxX,maxY and maxResultant

RD AD VC RPM Fmax Fx Fy
1 5 50 835 112 69 88
1 5 100 835 167 108 127
1 5 150 835 215 157 147
1 5 200 835 277 196 196
2 5 50 835 195 127 147
2 5 100 835 347 245 245
2 5 150 835 444 314 314
2 5 200 835 551 431 343
3 5 50 835 319 225 225
3 5 100 835 481 304 372
3 5 150 835 587 392 431
3 5 200 835 818 588 568


Having X and Y forces is more useful for our needs to size motors, whereas the Resultant force gives data on bending of the tool and load on spindle bearings.

Aluminium would be around 1/3 - 1/5 of these figures depending on spindle speed (faster = less force required but more spindle HP)

So a machine sized for 200N could take some 'serious' cuts in ally, say 3mm @ 150mm/min but would be limited to 0.2mm at 125mm/min in steel.

Taking the case of a simple XY gantry we need to accelerate the load to the cutting speed, say 125mm/min, against the inertia of the load, the friction of the slides, the friction and inertia of the screws and the internal inertia of the motor. For the sake of argument lets say acceleration is 1000mm/sec^2 and the load (work piece + table, etc or in this case the gantry and all on it) is 45kg on ballscrews (5mm pitch, 25mm dia, eff = 0.8) and good sliders (friction coeff = 0.1). We need 1.6Nm of torque to achieve this.

The motor torque is then reduced to keep the load moving at the desired speed, to approximately 0.5Nm. Next the work hits the cutter. We then have to increase torque to overcome the decelerating effect of the cutting load - this is calculated by likening it to friction and it needs 0.2Nm of torque to overcome this on top of the 0.5Nm to overcome the friction in the system, a total of 0.7Nm. A 200ozin stepper or servo will do this well and, by the same calculations, would easily handle the perpendicular load as there we are not moving the load only resisting the cutting force...

To maintain the same rate in steel with the same cutting depth would require a motor 50% bigger (plus beefier screws possibly but I've not done the calcs)

I was quite surprised how much torque is needed to spin up the screws - a 400mm long 25mm screw in steel weighs 1.5Kg!

This is a simplistic calculation and I'm sure there's more refinement that can be done, but at a first order is enough to be going on with!

[Al_The_Man]

Originally Posted by ftec Hi, (This question has probably been asked and answered in some other thread but I just can't find it).

See post #2 in this link.
http://cnczone.com/forums/showthread...=inertia+ratio
Al.

[ftec]

Originally Posted by Al_The_Man ... an older Emerson one. http://www.divshare.com/download/4412628-dac They almost all base their sizing around the industry custom of sizing the motor to load inertia as being max. 10:1 or less, the biggest influence on this is your accel/decel rate for a given load. ... Al.

Thanks for the link. I think I now have a resent version of the same program, now called CTSize by Control Technique. You may be able to download it from their web site after you have answered some questions, but I am not sure, got it from elsewhere.

http://www.emersonct.com/index.asp

The program accepts cutting force for load parameter and bases the calculation on accel/decel. It then accepts motors from the database based on the 10:1 ratio, all as you said.

Now using this program for sizing a CNC machine servo would be almost straightforward if it only were clear how to determine the acceleration needed when milling is in concern ???

IRVING:
Thanks for the enlightening examples and explanation. I'm a (retired) design engineer with very little experience of milling so I had no idea that the forces can be that small. I will try to make some inquiries myself of how machine shops determine cutting and spindle speeds etc. One source is probably the tool manufactures providing basic values for feed, spindle speed etc.

[irving2008]

Risto,

I found the same thing. Lots of programs/calculators on the web can tell you what happens in the unloaded state i.e. not cutting as regards acceleration, transit speed, etc. but thats basic newtonian physics and is only a few formulae. A lot of these programs are aimed at choosing a motor from the suppliers range but my issue is, having picked up some likely, but very cheap, steppers off fleabay, what I can achieve with them in the cutting state.

Mine are only 180inoz motors but with a heatsink and overdrven and/or a belt step-down should have enough torque to do what I need which is mill Ally, some light steel work and cut/drill PCBs. Traverse speeds for a home-unit aren't so much an issue, I dont have production deadlines to meet! I have a 1kW 3phase VFD drive for the spindle.

[ftec]

Originally Posted by irving2008 Risto, I found the same thing. Lots of programs/calculators on the web can tell you what happens in the unloaded state i.e. not cutting as regards acceleration, transit speed, etc. but thats basic newtonian physics and is only a few formulae. A lot of these programs are aimed at choosing a motor from the suppliers range ...

Yes, the programs are quite good for determining motors for say an XY assembly robot where you need to travel from point A to B, not really needing to know anything of the path the axis do while getting there. You just determine the type of speed curvature you want to have, trapezoidal etc.

With CNC on the other hand, I think we are talking about interpolation and this is where my knowledge ends. To use those programs for CNC motor sizing my intuition says you need to go to the "micro level" of the travel and speed curvatures to get the acceleration values or response times needed for the motor to stay in the position "window" and thus on the "cut track". I have seen what too small servos can be like in assembly robots and paper bag forming machines and the result is that you will never be able to get the PID parameters right. My past experiences may now be guiding me to the wrong direction but that is why I'm asking all these questions.

PS. I think I may have seen the light: the 10:1 ration may just be what is needed for a servo to be able to keep it's load in the "position window" ???

[irving2008]

I could be wrong, but I read the 10:1 thing as a way of ensuring reflected load inertia wasn't too small relative to motor interia. Clearly a directly coupled system wont have this issue, unless the motor is huge and the load is small (in which case you've wasted a lot of money for nothing!). But geardown with a toothed belt or gearbox to improve torqe and there is the possibility that the reflected inertia of the load drops into the 10:1 zone at which point it becomes harder for the control systems to resolve loading changes. I note that some suppliers quote 4:1 to 10:1 as the critical zone...

My motors have a rotor inertia of 260 g/cm2. The inertia of the ballscrew alone is 5 times that and my estimated 50kg load 11 times so the ratio is already 16:1... but with a small work item <1kg the ratio drops to 6:1. Do I have a problem? I dont know as yet...

[ftec]

Originally Posted by irving2008 I could be wrong, but I read the 10:1 thing as a way of ensuring reflected load inertia wasn't too small relative to motor interia.

Irving you must be wrong, the less load on the servo axis, the faster it can gain the position. These things are gradually coming back to me and I now remeber of a paper machine which had a very heavy roller cylinder (think of it as a flywhee) and the servo just couldn't run it /without crashing once in a while despite different PID parameters).

My motors have a rotor inertia of 260 g/cm2. The inertia of the ballscrew alone is 5 times that and my estimated 50kg load 11 times so the ratio is already 16:1... but with a small work item <1kg the ratio drops to 6:1. Do I have a problem? I dont know as yet...

Steppers are very much different from servos and have a high torque when speed is slow. I doubt that you can use the same rule of thumb for them.

[Al_The_Man]

Torque increases at the ratio of the reduction the inertia ratio reduces with the square of the reduction.
Servo torque curves generally show maximum torque at zero speed as does steppers.
The servo curve however drops off at a more gentle rate up to max rpm.
Al.