Stepper motor natural frequency
Stepper motor natural frequency
(OP)
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.
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.





RE: Stepper motor natural frequency
On the other hand, the formula doesn't look right either. Are you sure you shouldn't use the "spring constant" instead?
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: Stepper motor natural frequency
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I hope this helps.
RE: Stepper motor natural frequency
The equation I am using is:
Square root of (360degrees*Holding Torque)/(step angle*Inertia)
RE: Stepper motor natural frequency
0.3 degrees
------------
0.32 N*cm
The '200' doesn't belong in the equation.
Mike Halloran
Pembroke Pines, FL, USA
RE: Stepper motor natural frequency
RE: Stepper motor natural frequency
Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: Stepper motor natural frequency
I am trying to find the resonance.
RE: Stepper motor natural frequency
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
RE: Stepper motor natural frequency
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
RE: Stepper motor natural frequency
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
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: Stepper motor natural frequency
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:
ht
Unless you are considering the motor or the load to have close to infinite inertia?
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RE: Stepper motor natural frequency
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
RE: Stepper motor natural frequency
The units had me confused. I was referencing an old textbook and there answer was in steps/sec for the natural frequency.