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# Thread: Stepper moter sizing and power requirements

1. ## Stepper moter sizing and power requirements

Hello, Im design a CNC router for a college project, but im getting a bit stuck with the calculations.

I've done an experiment to try and calculate the force required to cut a groove though a sheet of MDF at a depth of 6mm with a woodworking router, this showed that with a mass of 8kg attached to the router with a piece of string would pull the router at said depth at a rate of 0.02m/s.

So, the question is, how can I use this information to calculate the power required to move that router in watts, instead of kg. And then use that to calculate the stepp

I'd really appreciate any information you can give me on this, I'm really struggling to get my head around it.

2. Specific cutting force of MDF is approx 6 N/mm^2. The force you applied is 8kg = 80N (approximating gravity to 10 and ignoring friction) with a 6mm DOC, so this suggests a 2mm diameter cutter which sounds small... what size cutter was it?

However the way to calculate required torque is shown below:

The required motor torque to drive a leadscrew assembly is the sum of three components: static friction torque, torque to resist the cutting forces and inertial torque, T = Tf + Tc + Ta

- Tf: Torque to move load against friction: The torque to move a certain load is a function of the lead and efficiency of the leadscrew assembly plus the friction of the linear bearings

where:
Load = N e.g. mass of gantry x 10 (for gravity)
u = friction coeff (oilite bearings is roughly 0.3-0.5, linear/ball bearings is 0.1)
e = efficiency of screw (0.8 for ball, 0.2 for acme or trapezoidal)

- Tc: Torque to resist cutting forces

Tc = CF x Lead/(2π x e) Nm
where:
CF = cutting force in N (this is your 80N)
e = efficiency of screw (0.8 for ball, 0.2 for acme or trapezoidal)

- Ta: Inertial torque:
Ta = (ls + IL + Im) x α Nm
where:
Is = The inertia of the screw in kgm^2 = π/32*density * length * diameter^4
(density in Kg/m^3 = 7850 for steel, length and diameter in metres)
Im = motor rotor inertia in kgm^2 (typically 2 - 5 x 10^-5kgm^2)
and

A typical acceleration for a small to medium machine is 2000rad/s^2 which allows it to follow a cutting path without under or overshoot.

Note: if geared/belt reduction (i.e. not 1:1 drive) then:
(Tc + Tf) is adjusted by the ratio
(Is + Il) is adjusted by the square of the ratio

Additional torque associated with driving and supporting the leadscrew must also be considered but can generally be accounted for by adding 5% assuming supporting ballraces and good quality couplers.

These torques are at the operating speed of the motor, not the holding torque. So once you have worked the torque T out you must work out the actual motor revs at the required cutting speed (depends on lead) and from the torque curves factor this back to the required holding torque (or use the fact that typically torque at cutting speed = 40% holding torque)

A working margin of 1.5 - 2 times is considered normal (so if 1Nm is the result of the above a 2Nm stepper would be selected)