I am building a lathe type machine to turn a tube, length 1.5m x 0.3m dia, weight 12Kg. The tube has a 30mm steel spindle supported at each end by a flange bearing. The drive is a chain drive using bicycle chain driving a bicycle chainwheel at one end of the tube. The reduction ratio is approx. 4:1. I want to turn the tube at about 1/4rpm so a motor speed of approx 1 rpm. The tube will be skimmed by a router, being pushed along a guide, to a depth of about 0.25mm as it turns. I would like to use an arduino controlled stepper motor as a tube drive, to give good low speed torque.
- Am I correct in saying for a speed of 1rpm with a stepper use a direct drive, or should i use a gearbox?
- What sort of size should I be looking at? I feel that a small NEMA 17 would suffice.
- Any other suggestions for motor drive at low speed? (I would like to keep costs down).
TIA
Hello @Robert72,
Based on this NEMA 17 motor specification (https://www.robotshop.com/en/12v-17a-667oz-in-nema-17-bipolar-stepper-motor.html), rated torque is 48 kg * cm which is equal to 4.70736 Nm.
Now let’s try and calculate torque based on your dimensions and T = I * a equation.
For I (mass moment inertia), I used this: https://www.omnicalculator.com/physics/mass-moment-of-inertia#moment-of-inertia-equation, and got I = 0.54 kg * m2.
For a (Angular acceleration), a = (W2-W1)/delta t. W1 is initial velocity which is 0 RPM and W2 is terminal velocity which you want to be 0.25RPM, and delta t is change in time, and let’s take it as 1 second.
This would give T = 0.54 * 0.25 = 0,135 Nm, which means that NEMA 17 motor would be enough for your application.
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Hello , thanks for the reply. The motor you linked to has a maximum speed of 34 rpm (0.567 cps). That was the point of Question 1 above. Also, the motor specs are ambiguous regarding gear ratio/step size. The step size is given as 0.018 deg, which gives a gear ratio of 100:1, for a motor step size of 1.8deg as per the sticker on the motor. But the gear ratio is given as 481.8/1 (991044:2057)?
Your calculation is a good start, but I get an I of 2.3175 kgm2, for a cylinder, mass 12kg, radius 0.15m, height 1.5m. This gives a T of 0.58 Nm. This is well under the 4.7Nm(?) of the gearbox motor, but does not include friction losses in the bearings, cutter resistance or drive losses (although I would imagine they all don’t come to much).
Sorry if this comes across as being a bit picky, but I thought i better reply so you can let me know if where I am going wrong.
Yes, motor’s max speed is 34RPM, so with a controller, you will be able to lower that speed.
I think gear ratio is 
.
Calculation is rather simplified. Maybe there is a mistake, but even with your calculation it shows that torque is well under 4.7Nm.
Anyway, when the tube reaches constant speed, the only torque that you need is one for friction losses and other losses, which shouldn’t be that big.
And it’s not picky, don’t worry 
60 rpm is 1 cps. 34 rpm is 0.567 cps. I need 1 cps. Gearbox ratio too high.
Yes sorry I just realised I wrote 1rpm but meant 1cps (I am used to using Hz, and meant 1 cps, or yes, as you say 60rpm). I think I just moved the goalposts halfway through the discussion. Apologies.
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