Gear reduction for Nema 23 4.2 A

[PeterJK]

I don't know if anyone mentioned it, but what most people do, when using rack and pinion with steppers, is to gear them to get between 10-25mm of travel per revolution of the stepper. The same goes for screw drive as well.

Thanks, Gerry.
So in my case gear reduction options look like this:

1:1= 69 mm/revolution
1:3= 23 mm/revolution
1:5= 13.8 mm/revolution
1:10= 6.9 mm/revolution

1:5 reduction should be ok, and will give me ample torque. Am I understanding this correctly?

[peteeng]

Hi Peter - The drive is current limited so if you connected two motors to one driver you would not cook it. I have done this on a couple of drivers to test (as cheap cncs do this) but the driver then gets weird feedback and you will either have midrange resonance or no high speed. You will need a 4 axis driver or 4 separate drivers.... I have run N17 motors on 1:5 reduction gearboxes and found that ratio to be a good spot for speed and torque. If your application needs fast rapids I'd go 1:3...Peter

[joeavaerage]

Hi,
3.8mH is medium inductance, low would be less than 2mH.

800 rpm with those steppers even with 70-80V driving them is optimistic. 500 rpm at 30% of holding torque is a more realistic figure.
Lets guess that the motor can manage 1Nm at 500 rpm, it could be less, it could be more, but not by much.

We are interested in the angular acceleration of the motor

dw²/dt²=Torque/J_total

where J_total is the total first moment of inertia of the system.

We have the rotational inertia of the motor = 800g.cm²
= 0.8 x10^-4 kg.m² (in MKS units)

Lets guess that the rotational inertia of the gearbox is the same as the motor, ie 0.8 x 10^-4 kg.m²

The most challenging accelerating mass is the gantry, 60kg. Given that there are two motors then the proportion to each motor is 30kg.
Lets equate linear kinetic energy to implied or effective rotational energy:

1/2 m.v²= 1/2. J_eff.w_pin²

where J_eff is the effective first moment of inertia of the gantry, or rather just one half of it, and w_pin is the angular velocity of the pinion.

But we have a relation between w_pin and v:
v= 0.069. w_pin/2.PI with 0.069 being the circumference of the pinon.
=0.01098 w_pin

Solving for J_eff:
J_eff= 1/2 m. v²/ (1/2. w_pin²)
= m . (0.01098²). w_pin²/w_pin²
= m . 1.205 x 10^-4
=30 x 1.205 x10^-4
=36.15 x 10^-4 kg.m²

Lets assume a gear reduction of 10:1 thus w_pin =w/10, so referred to the motor armature:

J_{eff, referred to armature}=J_eff/ 10²
= 0.3615 x 10^-4 kg.m²

J_total=( 0.8 + 0.8 + 0.36) x10^-4 kg.m²
=2 x 10^-4 kg.m²

Note that this is a critical discovery, the rotating components, the motor armature and the gearbox combined have a first moment of 1.6 X 10^-4 whereas the gantry has only 0.36 x 10^-4,
that is the rotating components dominate the inertia, NOT the weight of the gantry.

Computing the angular acceleration is now trivial

dw²/dt²= 1/ 2 x 10^-4
=5000 rad/s²

Converting to linear acceleration:

Accel =(5000 /10) / 2.PI x 0.069
=5.49 m/s² or 1/2g acceleration!!! Thats mega.

Note the acceleration is proportional to the square of the reduction ratio. So OP asks if 5:1 would work then the rotational inertia equivalent of the gantry goes up by (10/5)² =4.
Thus the gantries contribution goes from 0.36 x 10^-4 to 1.44 x 10^-4, and on a par with the armature and gearbox. The ;linear acceleration would drop to about
1/2, say 2.5m/s² or 1/4g, still pretty good.

Unless I've made an error I would say that 5:1 planetary gearboxes would be highly suitable.

Craig

[joeavaerage]

Hi,
reviewing my post I would say it makes sense. The whole point of using a gear reduction is to make it easier for a small motor to accelerate a heavy axis.
If you have a very great reduction, say 30:1 you might expect the that acceleration will be determined by the motor and gearbox because the reduction ensures that the weight of the gantry
has little to no effect.

At 10:1 reduction, which is what I based my calculation on, the inertia is dominated (80%) by the armature and gearbox with the gantry only adding 20%. At 5:1 the armature and gearbox
are 52% of the inertia with the gantry contributing 48% of the inertia.

If you were to decrease the reduction even further then the gantry weight would come to dominate the inertia equation and you would have to sacrifice acceleration, and high acceleration is
key to good tool path following.

With the data that we have I would conclude that reduction in the range of 5:1 to 10:1 is the 'sweet spot'.

To OP, do yourself a favor and get some high voltage drivers, low voltage cripples steppers. I would suggest 80VDC drivers and an 80VDC supply.

Craig

[PeterJK]

Craig - thanks for showing all of the calculations, that is very helpful.
Regarding the 80vdc power supply- the motors have a rated voltage of 24v (and I already have a 24V power supply)- wouldn't the motors burn out at 80v?

[joeavaerage]

Hi,

Regarding the 80vdc power supply- the motors have a rated voltage of 24v (and I already have a 24V power supply)- wouldn't the motors burn out at 80v?

No. The driver applies the voltage to the winding and them uses PWM to keep the current at its rated (4.2A) level. In practice the full voltage is applied for a few tenths of a millisecond or
less, thereafter the drive 'turns' the voltage way down. The 80VDC I recommend is to force the current through the winding quick smart. If you use 24V it will push the current through but it going to take
three-four times as long and you motor will stall at 100rpm. 24V is a joke. Give up on that, you need more, lots more.

You may have noticed that the specs for the motor says that the winding resistance is 0.9Ohm. Thus if you applied even 24VDC without a driver to limit the current then 24 / 0.9 =26.7A. Now 26.7A would burn
your motors out!!! Thus even at 24V you are relying on the driver to limit the current, and good drivers will keep the current steady at 4.2A even with 80VDC up its chuff.

Craig