Newbie- Load testing steppers on a Sherline system?

[David Campbell]

This is my first post in the CNCzone forums so please excuse the stupid questions... (I do realise that this post could belong in any number of forums)

Background: Recently I purchased a Sherline mill and installed a number stepper motors with a non-gecko driver and the system resonated like crazy above 5 revs/sec (which equates to 12 IPM). I finally bit the bullet and purchased a G540 and now achieving 89 IPM on the rapids (max out Mach3 at 60kHz rate, haven't tried a higher kernel speed yet...)

Getting decent speeds is one thing but under actual cutting load conditions without loosing a step is another matter, hence the issue of load testing (preferably without having to mount up a block of aluminium and making chips).

I have given this some thought and believe I might have a method which is:
a) reproduces the maximum machine load while cutting
b) easy to implement
c) cost effective

As some people might be aware the spindle motor of a Sherline machine is rated for 60 Watts continuous load. Assuming the spindle is running at 1000 RPM with a 1/2" bit the force on an axis can be calculated with a bit of maths:

Power (Watts) = torque (N.m) x rotational rate (radians/sec)
60 = torque (N.m) x (2 x Pi x 1000/60)
Torque = 0.573 N.m

Force (N) = torque (N.m) / cutter radius (m)
Force (N) = 0.573 / 0.00635 = 90.2 N (equiv to 9.2 kg)

Note: This is valid for either the X or Y axis

For load testing I would need to apply this force on the axis under test. Given the force is equivalent to the weight of 10kg, a bucket of water connected to the mill axis by a string over a pulley should do the trick. For a more sizable mill you may need to use an anvil for load testing.

It should be noted that the force is only applied in one direction so this test is somewhat artifical however it does have its advantages. When cycling an axis backwards and forwards it should only loose steps in the direction where the axis is loaded. Hence cycling the axis while lifting and lowering the load will accumulate missed steps in one direction only. A simple dial gauge should be able to pick this up.

Can anyone see any faults in the above test arrangement?

Now for the bit where I am struggling with the number crunching. When I attempt to convert the force into a stepper motor torque I end up with answers that appear to be way too small; using the force calculated in the above case the required torque is 1/30th of the motor holding torque of 2 N.m (270 oz-in). Either stepper motor torque drops dramatically when you get any decent speed or my steppers are seriously oversized.

David Campbell

PS: I am trying to figure out if the current 24 Volt supply for the G540 is enough or go for a higher voltage supply like 42 volts. The Gecko FAQ roughly states higher voltage => higher speed however there is very little in the way of guidelines about where you have gone off into the realm of an over-engineered solution.

EDIT: I thought the thumbs down icon was a question mark. Changed to "No Icon" instead.

[Fixittt]

I have the g540 running with a 24v xylotex PSu on a minitech mill. I only cut wax, but I can get this machine to fly!
If your wanting to load up the axis`s to simulate load, why noy just tap the gib adjustments in a little bit more to make the machine "TIGHTER"

For test cuts I have used MDF wood. Cheap to buy for test cutting.

I dont recommend running the sherline machine in the high rapids due to the metal on metal backlash nut design. It will wear faster.

[irving2008]

David,

Welcome. I concur with your thinking to a first aproximation as I have used this method myself. I am not sure though how you are calculating the stepper motor torque requirement, maybe you can expand on that a bit. The motor torque will depend on the leadscrew efficiency as well as load and motor inertia.. are you trying to calculate the static torque required to maintain constant feed speed or that required to accelerate the load to feed speed or what?

And motor torque drops off dramatically with speed, maybe only 15% of holding torque at 4000steps/sec. One way to help is to increase the drive voltage - a value of 32 x square root of motor inductance is considered a good place to start... but you need to keep 10 - 20% below the rated voltage of the drivers to allow for back-emf and transients as the max rated voltage for many drivers is the max static voltage of the chip set and not the actual max working voltage allowable... Gecko's should be good for 36 - 40v+ working as I understand it, but I have no personal experience of them.

[CarbideBob]

You're off to a good start. I like your plan of using a hanging weight to check the system. Nice to see someone actually doing the math behind their design.

You'll need the torque curves for you motor to see how much torque you have available at running speed. If you can't find exact charts for your motor/voltage take a look at some similar motors for a guesstimate. Holding torque is an advertising number and doesn't give you much info about system performance.

As mentioned above leadscrew efficiency must be taken into account.
You also need to add in the torque required to overcome the friction of the slide assembly and torque needed to overcome the gib force. Since you have the assembly already built you can check these numbers directly by measuring the force required to rotate the screw.

The big requirement is usually the torque needed to accelerate the system. Remember to add the inertia of the screw, coupling, and motor rotor along with the inertia of the slide and the max workpiece weight.

Bob

[David Campbell]

Fixitt: Point taken with metal-on-metal wear

Irving: The problem is not so much whether the G540 can take the voltage as is there any benefit (the filter capacitors start getting pricy as the voltage increases)

CarbideBob: Thanks for the feedback & suggestions, I did some calculations and the bulk of the torque requirements appear to come from the leadscrew / motor rotational intertia.

Warning: Calculations follow...

Motor rotational intertia: 480 g.cm²
Leadscrew rotation interia: 687 g.cm² (includes coupler and handwheel)
Total rotational intertia: 1167 g.cm²

Screw ratio: 1 mm = 2 pi radians
Assuming an acceleration rate of 100 mm/sec² => 628.3 rad/sec²

Torque due to rotational intertia = Inertial x rotational acceleration
Torque = 1167x10^-7 kg.m² x 628.3 rad/sec² = 0.0733 N.m
[10^-7 is the conversion from g.cm² to kg.m²]

Force due to acceleration of load = Mass x acceleration
Force = 16.3 kg * 0.1 m/sec² = 1.63 N

Converting the leadscrew axial force back to a torque is complex, see http://www.engineersedge.com/gears/s...alculation.htm for details (includes friction losses).

For the Sherline leadscrew the relationship is approximately:
Torque (N.m) = Axial force (N) x 0.001
Torque = 0.00163 N.m

16.3 kgs is the weight of the entire mill, this is way larger than anything that I would intend to cut but I am using this to prove a point. Even for this "worst case" scenario, the rotational intertia is approx 45 times larger than the acceleration of the load.

Factoring in the load due to cutting (see original post) the total torque numbers are as follows for an acceleration rate of 100 mm/sec²

Cutting force: 0.090 N.m (12.8 oz-in)
Rotational intertia: 0.0733 N.m (10.4 oz-in)
Mass intertia: 0.0016 N.m (0.2 oz-in)

Total torque: 0.165 N.m (23.4 oz-in)

From various stepper motor charts, at about 4000 steps/sec you should only expect 15 to 20% of the holding torque (most of the curves were for a 24 volt system). In the above case the total torque is 8% of the holding torque so there is a nice safety margin but not hideously excessive.

That was quiet a bit of an eye opener that the rotational inertia is one of the largest torque contributors

David Campbell

[Mariss Freimanis]

David,

Nice calculations and I always appreciate someone taking an analytical approach rather than an empirical, seat of the pants solution.

1) Step motors are essentially 'constant power' motors. That means the product of torque times RPM is constant at higher speeds, meaning torque is the inverse (1/x) of RPM.

2) What matters is power because it's what gets things done. Power is torque times RPM. Since motor power is a constant at higher speeds, all you have to do is measure power accurately at a single speed to computationally derive what your torque will be at any speed. How do you do that?

2a) What you need is a paper towel, a pair of channel-lock pliers and a multimeter.

2b) Dismount the motor. Take a paper towel and fold it over as many ways as you need to until it is about 2" long, 1" wide and at least 0.25" thick (50mm by 25mm by 13mm).

2c) Fold the 2" (50mm) length around your motor shaft ('U' shaped). Wet the paper towel very slightly if more than 50W is expected from the motor.

2d) Place your channel-lock pliers around the folded towel on your motor shaft. The pliers will be the brake calipers, the towel will be the brake 'shoe'. The towel will dissipate the motor's mechanical power (by boiling the water used to wet it it) while protecting the shaft from being gouged by the pliers.

2e) Connect only one motor (the test motor) to the G540. Set your multimeter to 'DC Amps' and put it in series with your power supply to the G540.

2f) Run your test motor up to 1,000 or so RPM. Do it without the paper towel on the shaft. Measure the power supply current and write it down as 'Amps no-load'.

2g) Place the towel and channel-lock pliers on the motor shaft. Very gradually apply pressure to the plier handles; watch the current as you do, it will increase. Keep increasing pressure while watching the current until the motor stalls. Catch the reading at the instant of stall. This may take several tries until you can sneak up to that point accurately.:-) Mark that current as 'Amps stall'.

2h) Subtract 'Amps no-load' from 'Amps stall'. Multiply the result by your power supply voltage; the result is your motor mechanical power output in Watts (Watts = VDC * (Istall - Ino-load)).

2i) The results from (2h) allows you to compute torque at any speed where the result is less than the holding torque (low-speed torque) of the motor. The general equation is:

in-oz = Watts * 1351 / RPM. If you prefer Nm, divide in-oz by 141 to get torque in Nm (Nm = 9.58 * W / RPM). If the test is carefully done, the accuracy will be within +/-5% of what you would get from a dynamometer.

Note: We use a custom-built 0 to 1,000 in-oz dyno that has +/-0.3% linearity, +/-2 in-oz offset error accuracy. This method has been repeatedly tested against it to verify the above +/-5% claim.

Mariss

[David Campbell]

Mariss,

I will look into running the recommended tests tommorow, probably at several different speeds to get a feel of the "linearity" of the torque-speed curve (plot on a log-log chart it should end up being a nice linear line). The 1000 RPM tends to be the corner speed for a number of stepper motors and for the test to provide meaningful numbers it should be run a little higher.

I suspect wrapping a couple of layers of electrical tape over the shaft to remove the sharp corners of the shaft flat wouldn't go a stray. I am imagining that the shaft flats could easily tear up the paper towel.

Question: You state that a stepper motor is a 'constant power device', is this a 'constant power' for a fixed drive supply voltage?
Edit: I found the answer elsewhere - power is proportional to V / sqrt(L)

David Campbell

PS: My background is chemical engineering so the calculations don't bother me - it is a matter of digging out the text books for the appropriate mechanical engineering formulas. The CNC machine is for hobby purposes.

[David Campbell]

Mariss,

Load testing a single stepper motor (as per your instructions) went well (except that the electrical tape I added effectively melted due to the friction) and got some useful information.

I noticed that the stall current when divided by phase current was 0.61, which happens to be:
- the recommended ratio for a G540 (total phase current of all motors x 0.6)
- the DC:AC current ratio of a full bridge rectifier (~0.62)

Then I realised that the stepper motor drive can be thought of as a variable speed inverter... Perhaps I'm reading too much into the figures and it is merely co-incidence.

A quick question about iron losses. I notice that iron losses increase with voltage (page 13 of Geckodrives "Step Motor Basics"), is this due to voltage directly? or due to higher step pulses that can be achieved due to higher supply voltages?

David Campbell

[Mariss Freimanis]

1) You may not be turning fast enough or are not sneaking up on the stall-point accurately. Some square motors can approach a 1:1 ratio. A shaft with a flat is extremely difficult to load accurately. Use a shaft coupling.

2) Think of a step motor as a fixed RPM motor driving an infinitely variable gearbox. The step pulse frequency sets the gearbox reduction ratio.

3) Iron losses are due to I^2*R dissipation of currents (called eddy currents) induced by changing magnetic fields. Increase the supply voltage and you proportionally increase these currents. Dissipation is the square of current so theses losses increase as the square of the supply voltage.

Mariss

[David Campbell]

I'll try a shaft coupling and see if I can get a better result.

Your comment about iron losses took a while for it to sink into my thick skull. Initially I thought the whole idea of PWM is to have a near constant phase current with minimal ripple.

After several hours of head scratching it dawned on me that the current, and therefore the magnetic field, is constantly increasing/decreasing by about 10,000 amps/sec (Vsupply/L). As this is happening at 20kHz the net current ripple is only about 5% of the rated current. This "rate of change" in phase current is inducing a voltage in the laminations which results in the eddy currents (P_loss = V_induced^2 / R_lamination).

What would be interesting to see is what difference the phase current has on the core losses. I suspect running at half phase current would reduce the copper losses however there would be no change in core losses.

The only way to reduce the effect core losses would be to vary the supply voltage as a function of step rate. In theory you could create a small switch mode power supply with the reference voltage adjusted by a circuit monitoring the step pulse signal. However for the cost of this you might as well buy a decent stepper motor.

David Campbell

[Mariss Freimanis]

You are correct; copper losses and iron losses are entirely independent. Copper losses are independent of microstep location because the currents are sine and cosine making dissipation sin^2 + cos^2.

Iron losses are eddy currents induced in the stator laminations by varying magnetic flux. This flux variation results from the turning rotor magnet and di/dt effects if the drive is a switching type.

You can demonstrate the eddy current losses caused by a turning rotor buy chucking an open winding motor up in a drill-press and turning it at a high speed. The motor will become very hot even though no current is flowing in its windings. The power dissipated in the motor is equal to its detent torque times RPM.

Mariss

[Crevice Reamer]

Hi David. Welcome to the Zone!

I applaud your science. But this is a tiny SHERLINE--Not a Haas. A sherline powering a 1/2" end mill can only take very light cuts. You are NEVER going to be able to just hog out material with it.

Light and easy does it.

CR.