Iirc, there was an article in MEW about adding a 3 ph motor and vfd to a lathe, cant remember if it was the WM250……

I'm converting my SPG lathe ( I think it may be the equivalent of the 290 but with a brushed motor…) to 3 phase with a 1.5kw motor ( going for a 2 pole job, the 4 pole 1400 rpm version only comes in a 90 frame size…tooooooo big enough ) and vfd, but undecided on choice of inverter….

I chose a 2 pole motor for 2 reasons, at low spindle speed, the motor will still be turning resonably quick to avoid any overheating issue, and secondly, the 80 frame size motor has the same footprint…I've turned up a replacement pulley to suit the motor shaft…14mm rather than mod the orig…if at some point I want to revert back, I can…

Edited By John Rudd on 27/09/2015 16:41:24

Mark,

You're correct, the ratios will be the same.

If you use a 2 pole motor, its "native" speed is 2800rpm at 50Hz. If the VFD is set to 82Hz the motor will run at 4600rpm.

Above 50hz, the motor will give full torque (same as at 50hz), but below 50Hz the power will be the same as at 50Hz, but the torque will be reduced (power = torquexRPM).

Edited By Gary Wooding on 28/09/2015 15:41:26

Let's get this straight, assuming we're dealing with an induction motor with a base speed at a frequency of 50Hz. Base speed generally means where the motor is running at it's rated voltage, and hence rated current and up to rated power.

Above 50Hz, at the rated voltage, there is less current, as the motor is turning faster and therefore there is more back emf opposing the current. So the current, and torque, go down but since the angular velocity has gone up the power stays roughly constant. So, above base speed you get constant power.

Below 50Hz more current can be delivered at the rated voltage. However, to protect the motor the current is normally limited to the rated current. Since the current is constant the torque is constant, but the power decreases as the angular velocity reduces. So below base speed you get constant torque.

To be precise the above equation should be:

power (Watts) = torque (Nm) x angular velocity (radians/second)

Andrew

I agree with Andrew. If Gary's statement was correct it would imply infinite torque at zero speed.

Les.

Same here, only used it once when new to see what it was like and never again.

Still not sure that running the motor at 50rpm with the 1:1 ratio would not give overheating problems

Mark,

I run a 1.1kW 80 frame 4 pole motor on my inverter converted WM280.

I use taperlock z section pulleys of size 71mm (motor) to 106mm (Spindle)

This motor power and belt /pulley setup should ideally be run with twin belts, a larger motor pulley would reduce this need, in practice one belt is fine most of the time, I couldn't be bothered to make a custom mount for the motor to facilitate lining up the pulleys to run 2 belts.

The pulleys give a ratio of circa 0.67, and in practice give a speed range of 75-2000 Rpm.

If I were to spec this up again I would go for a ratio nearer 0.5, which would give a slightly more usable low speed with a top end still at 1500.

The current setup with just 1 belt means the belt slips before the motor stalls, which is fine for my somewhat beginner level experience.

Haven't a clue, as I haven't done any turning at that size yet, though the motor speed is about 110rpm at a spindle speed of 75.

I originally intended to get two pairs of pulleys, for high and low, but i went down the middle just to get the lathe up and running, I haven't found the need to change yet. I must get around to measuring up again for the largest size that'll fit under the guard so I can make up a low range set, I think it will be needed eventually.