r/diycnc • u/Horloscarlita • 2h ago
DIY CNC Router: is the math mathing?
I have a lot of cabinet building in my future, so I’m looking into building a large-format CNC router that can handle full sheets of plywood. I’m thinking of a metal construction (square tubing and ~4 mm laser-cut plates) similar in design concept to this CNC plasma cutter: https://www.youtube.com/watch?v=2jZwh6mvMhQ
What I’m trying to figure out is the motion system for the X and Y axes, as well as a target feed rate. So, here is my idea:
I already have, and want to reuse (from a failed project from my high-school days):
- 2.2 kW water-cooled spindle with VFD
- 3x 23HS9442 NEMA 23 steppers(2.8 Nm), DM542A drivers at 36V
According to https://cncshop.com/pages/calculating-feeds-and-speeds I figured that if I want to cut plywood with a two-flute, 6 mm bit, I need a feed rate of 5,000–6,000 mm/min. If I want to cut a pocket with a two-flute, 10 mm bit, I need 14,000–16,000 mm/min.
For a CNC of this size and application, I’m considering an HTD5M, 15 mm or 20 mm timing belt, with a pulley coupled to a planetary gearbox.
Even though I cannot find a datasheet for my stepper motors, after looking at torque curves for other similar-sized steppers, I concluded that 250 rpm is the speed at which the torque starts to drop off.
With all this out of the way, let me explain my calculations. The first table shows the output torque, calculated as follows:
Output torque = stepper rated holding torque × fraction of holding torque at operating speed × gearbox ratio × gearbox efficiency
I have assumed a fraction of 0.7 for the running torque and a gearbox efficiency of 0.9.
For example, for a 3:1 gearbox:
2.8 N·m × 0.7 × 3 × 0.9 = 5.292 N·m
The second table shows the axis linear backlash based on the gearbox backlash and the pulley pitch diameter:
Backlash = pitch diameter × pi × (gearbox backlash / 360)
For an 18-tooth pulley with a 28.65 mm pitch diameter and 0.5° of gearbox backlash:
28.65 mm × 3.14 × (0.5 / 360) = 0.125 mm
The third table shows the feed rate, in mm/min, for a stepper running at 250 rpm, depending on the gearbox ratio and the number of pulley teeth:
Feed rate = (number of teeth × tooth spacing / gearbox ratio) × stepper rpm
For an 18-tooth pulley and a 3:1 gearbox:
(18 × 5 / 3) × 250 = 7,500 mm/min
The final table—which I will probably get some hate for—is the equivalent mass being pulled on the gantry in kilograms. The gantry has two stepper motors. I calculated this to get some feeling for the force generated by the motors when moving the gantry:
Equivalent mass = torque / pulley radius / gravitational acceleration × 2
For a 24-tooth pulley and a 3:1 gearbox:
5.292 / (0.0191 * 9.8) * 2 =56.5445 Kg
Okay, the math session is over. What do you think? Do you agree with the logic? What combination of pulley size and gearbox ratio would you choose? A 28T pully and 3:1gearbox gives a speed of 11666mm/min and a “oomph” equivalent to 48.4Kg – I think that is a good compromise.
PS: I will be posting this in multiple CNC groups.