Stepper motor natural frequency 3

[eulerpi]

I am trying to calculate the natural frequency of a stepper motor.

Here is my work so far.

The motor is rotated .3 degrees with a torque of .32 N*cm.
Inertia = 48 g*cm^2

200 step/rev
1.8 degrees per step

I am using the equation frequency = sqroot of (200*.32)/48
Which = 1.1547 steps/sec

I am new to working with stepper motors so I am not sure if this is correct.

I expected to see a higher value around 155 steps/sec.

Thanks for any help or advice.

 

[Skogsgurra]

Without having checked that formula, I can say that you probably have missed the factor 100 that you get when using cgs instead of SI.

On the other hand, the formula doesn't look right either. Are you sure you shouldn't use the "spring constant" instead?

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Clyde38]

If by "natural frequency of a step motor" you mean that you are trying to calculate the resonant frequency, take a look at this site and read about resonance.

http://www.cs.uiowa.edu/~jones/step/physics.html#dynamics

I hope this helps.

 

[eulerpi]

I agree it appears that I am off by a factor of 100, but I am not sure why.

The equation I am using is:

Square root of (360degrees*Holding Torque)/(step angle*Inertia)

 

[MikeHalloran]

Yeah, the 'spring constant' of the magnetic detent would be:

0.3 degrees
------------
0.32 N*cm

The '200' doesn't belong in the equation.




Mike Halloran
Pembroke Pines, FL, USA

 

[eulerpi]

What equation should I be using?

 

[Skogsgurra]

I think it would be better if you understand the problem instead of plugging dimensionally incompatible units into a (possibly wrong) formula.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[eulerpi]

I know that the square root of the torque stiffness divided by the Inertia equals the Frequency.

I am trying to find the resonance.

 

[cswilson]

There is a very basic confusion in your problem statement. A frequency is fundamentally in units of inverse time (1/sec). A "step", which is a measure of angular distance, has nothing to do with it.

The fundamental equation for resonant frequency of an angular system with some kind of "spring" action is:

wn = sqrt (k / J)

where:
wn (omega-sub-n) is the natural frequency;

k is the angular spring constant (torque per unit of angular offset);

J is the moment of inertia of the system

Note that you must be VERY careful with units. First, you have a hybrid of MKS and CGS units, so you must convert for consistency.

Second, you must understand the difference between your physical (mechanical) angle units and the "virtual" angles that express a part of a time cycle.

That said, let's start.

I am assuming that your problem statement indicates that if you perturb your motor 0.3 degrees mechanical (1/6 of a step) from equilibrium, the motor (with a given current applied) will generate 0.32 N-cm of countervailing torque. This means that your spring constant can be expressed (in inconsistent units) as:

k = 0.32 N-cm / 0.3 deg(mech)

To get this into constistent units, we first note that:

1 N = 1 kg-m/s^2 = 1000 g-m/s^2 = 100000 g-cm/s^2

So 0.32 N-cm = 32000 g-cm^2/s^2

Next we convert the angle:

0.3 deg(mech) * Pi/180 rad(mech)/deg(mech) = 0.00524 rad(mech)

So our spring constant in consistent units is:

k = 32000/0.00524 = 6111550 g-cm^2/s^2

Our resonant frequency is:

wn = sqrt[(6111550 g-cm^2/s^2) / (48 g-cm^2)]

= sqrt[127323 / s^2]

= 356.8 /sec [aka 357 rad(time)/sec]

= 56.8 Hertz

I'm not sure what you meant when you said you were expecting a frequency of 155 steps per second -- that is an angular velocity, not an oscillatory frequency.

Curt Wilson
Delta Tau Data Systems

 

[MikeHalloran]

The natural frequency of a bare unpowered step motor is actually not of much interest. In service, the dynamics are also affected by the inertia and friction of the driven load, and ... to a greater degree than in other types of motors ... by the dynamics of the electrical drive circuits.

If you're interested in the mechanical dynamics at a theoretical level, see the PhD thesis of R.K. Gauthier at University of New Hampshire, and other papers by he and C.K. Taft.

For a more practical approach, buy A.C. Leenhouts' book, and/or snag some of the stuff he wrote for Superior Electric's application notes.





Mike Halloran
Pembroke Pines, FL, USA

 

[Skogsgurra]

Yes, correct. I insist on using SI. That gives:

k = 0.0032 Nm/0.0052 rad = 0.61 Nm/rad
I = 0.048 kg * 0.0001 m^2 = .0000048 kgm^2

w = sqrt(.61/.0000048) = 357 rad/s; f = 357/6.28 = 57 Hz

Which is the same result.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[electricpete]

I'm not familiar with what you're doing.

In general, most rotating systems require two inertias to determine a torsional resonant frequency. For example I have prepared a spreadsheet for torsional resonant frequency in English units here:

http://home.houston.rr.com/electricpete/eng-tips/Torsional.xls

Unless you are considering the motor or the load to have close to infinite inertia?


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[cswilson]

pete:

I don't believe this concerns a resonant frequency of coupling. Eulerpi asked about "natural frequency", which is usually different.

I think the "spring action" in this case is the restorative force of the stepper motor itself about its no-load position. Whether this is due to unpowered cogging torque or powered Lorenz torque (or a combination), I don't know.

Both skogs' and my calculations dealt with an unloaded motor rotor. Adding a load, even with a perfectly stiff coupling, will increase the inertia and therefore decrease the natural frequency. If the coupling has substantial compliance (limited stiffness), it will add a second frequency of interest -- the resonant frequency you are thinking of.

Curt Wilson
Delta Tau Data Systems

 

[eulerpi]

Thank you ALL for the help.

The units had me confused. I was referencing an old textbook and there answer was in steps/sec for the natural frequency.