Its because a reduction of spindle speed is done electronically and not by a step down gear box that automatically increases the spindle torque.

Frank

I have the Warco WM280VF which is very close to yours and don't have problems at that diameter but what do you call "modest" Its only when doing 9-10" stuff that it can be slowed buy I just run a bit faster and take slightly shallower cuts 0.025" on a 10" CI flywheel is no problem. Mine will also part 3" steel and 4" CI OK.

Firstly make sure you have it in the slower of the two speed ranges which will give a better mechanical reduction from motor to spindle. Then check its not th ebelt slipping.

Edited By JasonB on 11/04/2016 07:42:09

Edited By JasonB on 11/04/2016 07:45:07

Does sound like something is amiss, did you check it was in the lower speed range? also have a look at the brushes on the motor

This steel fabrication has a bit of 10mm black flat bar on the top which is not the nicest steel to machine. When it gets past the intermittant cut the dia is 80mm, and I think it was 0.020" depth of cut or 0.5mm

And a bit of fun the other day, 1" dia EN1A being reduced to 1.2" dia in one pass so 0.250" or 6.35mm depth of cut. I don't usually go that deep.

Even more impressive is Jason reducing a 1" diameter to 1.2" diameter…arry potter got nown on Ja!!

And if you need to buy a spare belt the I can recommend Beeline Engineering in Milton Keynes…no connection et-al…

I bought a spare for my SPG lathe cheaper then anywhere else.

If you want decent torque and no complications in a machine then it must include backgear IMO

Jon,

If you halve the torque and the rpm then the power is 1/4 of the full speed output.

The VFD system has to lower the applied voltage to the windings to keep the stator and rotor currents below a safe maximum. The main thing opposing the ac is the reactance (X) of the inductive winding on the stator. The stator resistance is normally quite low but the reactance is linear with frequency. X = 2Pi x F x L where F is the frequency in Hertz and L is the inductance in Henry's. As the frequency drops the current would go up as X is reduced, so the applied voltage is reduced to compensate. Torque to a first order is determined by the current flowing in the windings, as it generates the magnetic flux. At high frequency the reactance is now high and so to get sufficient current flowing the applied voltage has to be increased.

The rotor essentially is a massive shorted turn which when the magnetic field is applied by the stator has a very high current flowing in it. It behaves like a normal transformer where the stator is the primary winding and the rotor is the secondary winding. The rotor effectively only has 1-turn and rotor current is essentially the turns ratio between the primary and secondary, just like a static transformer. So if the stator (primary winding) has, say 100-turns then the secondary current in the rotor will be 100-times the stator current in simplified terms. Because of this the rotor current can become high enough to cause overheating.

The rotor induced magnetic flux works with the rotating flux supplied by the stator windings to produce the torque, but only when "slip" occurs, which is always the case with ac motors. If the rotor did manage to get up enough speed to equal the applied frequency then slip drops to zero and so does the torque, hence it slows down until enough slip occurs to produce the torque required to drive the load.

Below a certain rpm the cooling effect is decreased and the rotor begins to overheat. NEMA recommends that a motor should not be operated below 1/3rd of its normal speed to ensure sufficient cooling is available from the internal fan. For applications where very slow speeds are required then an external fan provided by another motor is required.

Hi Jason,

Then why was Jon describing an ac motor? I am confused!