[creep_pea]

Hi

I've just bought a Smallpiece Cromwell S.S. & S.C. 3.5" x 19" Super-Precision Model S.800 Lathe and I need some help to wire it all back up.
The spindle was orignally driven by a "Ward-Leonard" electrical system that comprised a 3-phase motor attached to a DC generator that in turn supplied current for a DC motor controlled by an early form of electronic variable-speed drive but at some point in its life the 3 phase motor has been replaced with a single phase motor.
I get the feeling it may never have run like this and may not even be possible to do.

The attached drawing shows how it is wired at the moment so if it is possible to do it like this where do the two wires shown in red going into the AC motor go?

Cheers

Chris

[Al_The_Man]

Hi

The attached drawing shows how it is wired at the moment so if it is possible to do it like this where do the two wires shown in red going into the AC motor go?

Boy, You sure that does'nt date from the time of Cromwell
I did not think there was any of those in existance still.
The diagram does not look right, the AC motor should drive a DC generator with a variable field control which in turn feeds a DC motor with fixed field.
see diag.
Al.

[gar]

050817-0827 EST USA

creep_pea:

Superficially this circuit makes no sense.

If you interchange the field and aramature on both the dc generator and the dc motor, then it starts to look better. The ac motor would not normally be drawn with the implication of brushes and a commutator.

My guess is that the 512 ohm resistor with the arrowed line thru it is a potentiometer rather than a variable resistor. The potentiometer feeding the field of a dc generator makes sense because this provides a means of varing field excition from zero to maximum, for a generator output voltage from 0 to maximum.

Also full field excitation is supplied from the bridge rectifier to the motor at all times. (This is on the assumption that the motor and generator are both drawn wrong.)

The two terminals on the ac motor --- no idea at this time.

The labels coil and 20 ohm probably need to be interchanged. It appears this is a relay, but as drawn would probably oscillate.

There is nothing drawn to detect loss of field excitation to the motor field winding. A field excitated dc motor will runaway if field excitation is lost.

.

[creep_pea]

Sorry electrical numpty here.

All the motors/generators are disconnected at the moment I just guessed were they went on my circuit diagram whops!

I don't know the difference between a potentiometer and a variable resistor so you are more than likely correct, I'll post some pics later today to be sure.

Thanks very much for the help

Chris

[Al_The_Man]

Going by the size of the lathe, I would assume the motors are not that large?
You might want to look at picking up a more modern DC motor controller and getting rid of the AC motor DC generator. Ones for wound field motors, can often be had cheap surplus.
Al.

[gar]

050817-1015 EST USA

creep_pea: Chris:

As a solution to your problem Al's idea is a good one.

A potentiometer is a voltage divider, a fixed resistor with a variable tap contact. A variable resistor, rheostat, is a potentiometer with one end of the fixed resistance unused. The difference in names relates to functionality and useage.

If you connect the outer terminals of a linear potentiometer across a constant voltage source, and apply no load to the sliding contact and measure the voltage of the slider relative to one end of the fixed resistor, then the voltage will be proportional to the position of the slider. Adjustable from 0 to the source voltage. Potentiometers are used as volume controls in radios, but are usually made with a log instead of a linear curve.

Put a rheostat between a voltage source and a high impedance voltmeter and adjust the slider and there is no voltage change. In this case to get a voltage change there must be a finite load resistance, and output voltage will range from full source voltage to something less but not zero. The output voltage will not be linear with slider position. The output voltage is Vo/Vin = Rl /( Rv + Rl ). Where Vo is output voltage, Vi is the input , Rl is the fixed load, and Rv is the variable resistor.

If the load is the variable resistor, then Vo/Vi = Rv / ( Rf + Rv ) . Where Rf is the fixed series resistance and Rv is the variable and load resistance.

.

[creep_pea]

Here are a few pics for you to look at.

[Al_The_Man]

I see somone has put some 'modern' silicon rectifiers on it
In light of the age & condition, another solution would be to pick up a 240 3ph motor and run a VFD off of single phase.
Al.

[creep_pea]

Al

Thanks for the idea might pursue this if I can't get it working as it is, but for now I would just like to wire up what allready there (if possible) and see how it goes.

Cheers

Chris

[Al_The_Man]

As GAR mentioned, there should really be field-loss detection on the motor, if you lost the field control for any reason the motor can wind up fast to dangerous rpm's, some time causing disintegration of the rotor.
Al.

[gar]

050817-1044 EST USA

Chris:

Your bottom right picture is the potentiometer. It is clear this is used to adjust the generator field current from 0 to some maximum value. This provides from zero to a maximum voltage out of the generator to supply the motor armature.

At fixed field excitation a dc motor speed is proportional to armature voltage.

If you do not follow Al's suggestion, then redraw the schematic as I previously described. Your photos do not provide information on the two wires from the dc area to the ac motor.

But Al's idea is going to get you further, faster, and better.

Visited your county many years ago. We anchored at Edinburgh and I took a tour of London via train. Meat was in short supply in your country at that time so we had hamburger buns with greens for our meal on the train. As we came into Edinburgh the fog was hanging low, but the castle was visible as if floating in the sky. In London we toured all the special points.

.

[wotje]

This AC motor is probebly a single face asynchrom motor. Which is reverseble in direction.
That means it needs only two connections to line input. And the other connections should be made as on the schematic. Or you must have a direction switch outside. To make this motor run.
Otherwise there are no connections to other wires which are comming from outside, such as you draw in your drawing.
The connection in your drawing between de AC motor and the DC generator makes no sence. If you run it, Ithink it go's *!@#$%^^&** hell what a fire.

Greetings

[gar]

050817-1125 ESY USA

Chris:

First: Get your ac motor to run without the two wires that go over to the dc area.

Second: Mechanically couple the generator to the motor, with no power to the generator field.

You need a VOM (volt ohm millameter).

Third: Determine for sure which two terminals on the generator are the field. These will show a resistance of 50 to 500 ohms by my guess. The armature will be much lower.

Fourth: Run the motor and generator. You should see some voltage from the armature, and none from the field. The armature voltage comes from the residual magnetic field in the core.

See if you can get this far.

.

[creep_pea]

Gar

Here is another version of my circuit diagram with fuse box, been looking at armature and field wiring and I had definately got them the wrong way round it even tells you which is which on the fuse box.

Also I noticed I've got the switch and coil in the wrong place on my first circuit diagram. The coil is shown in the picture below I originaly put a 20 ohm resistor on the diagram to show the resistance of the coil.

[Al_The_Man]

I agree about the initial hook-up, I cannot see why you would have any connections coming from the AC motor, it looks like a standard split-phase capacitor start motor. The only other AC feed you should require is to the Bridge rectifiers.
Al.

[gar]

050817-1317 EST USA

creep_pea:

The new circuit is looking better.

You will need to find out if there is a switch contact on the motor that is normally open when the motor is stopped and closes near full motor rpm. That would make sense relative to your new diagram.

Is the item you label as switch really a normally open contact on a relay with a coil of 20 ohms?

A coil with a parallel switch contact does not make sense in this circuit. 20 ohms seems high for a coil to go in series with the motor field, and too low to go in parallel. Further a parallel sensing of motor field is not good. Current sensing is best. The logic of the switch (relay contact) is correct for using in detection of loss of field because this would open the motor armature circuit.

My best guess is that the coil you have drawn in parallel with contact labeled switch should be wired in series with the motor field winding. Check this coil resistance, and the resistance of the motor field and report these back to us.

.

[creep_pea]

Gar

Right I've got the single phase AC motor running and determined the terminals for the fields/armatures on both the DC generator and motor. Will try and get AC motor and generator connected later today.

I'm fairly sure the switch is a switch and not a relay, it's the bottom conntact on the Dewhurst drum-type reversing switch and the coil is definatly wired parallel to it.

[gar]

050818-0626 EST USA

creep_pea:

Since the switch is on the reversing switch tell us when it is closed and open relative to the position of the reversing switch.

What is this 20 ohm coil associated with, or is it simply a 20 ohm power resistor. My guess is it is a 20 ohm resistor to limit current during the transfer of the forward-reverse switch from one state to another. Thus, in forward or reverse the switch would be closed, but in between these two states it would be open. Maybe also a time delay to close.

At some time you need to get a current sense relay in the dc motor field line to shut off armature voltage on loss of field current.

If I understand you photos it looks like your field coils are about 300 ohms. Thus at 200 v your field current will be in the range of 0.7 amps. If you used a 12 v battery charger to excite your generator field you would probably get about 10 to 30 volts out of the generator.

Any further information on those dc leads to the ac motor?

.

[creep_pea]

My guess is it is a 20 ohm resistor to limit current during the transfer of the forward-reverse switch from one state to another. Thus, in forward or reverse the switch would be closed, but in between these two states it would be open. Maybe also a time delay to close.

I think you've hit the nail on the head there, the switch is open when off and closed in either forward or reverse

[creep_pea]

You will need to find out if there is a switch contact on the motor that is normally open when the motor is stopped and closes near full motor rpm. That would make sense relative to your new diagram.

How would I find this out? I think I am right in thinking the motor is a capacitor start/run motor as it has 2 capacitors, so it would have a centrifugal switch to change from the large capacitor on to the smaller capacitor or am I way off the case?

Cheers

Chris