An electro-mechanical problem to solve - vintage chart recorder

I'm not getting very far figuring out how my vintage chart recorder paper-feed drive mechanism works.

If you know your 1Ph synchronous motors, including hysteresis synchronous motors, or synchronous motors with capacitor permanently in circuit, then you may have the edge.

Here is the paper-feed mechanism and the circuit that I copied:

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I'm bamboozled by how it works myself.

All ideas taken into consideration.

Thanks if you solve it's operation.

Reply to
Richard
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M1>---------------------------------SHAFT

The thing is what does M1 do?

Its moving too slow and in the wrong direction turn the SHAFT at the correct SLOW speed.

And you would think M2 might not be able to go slow enough to turn the SHAFT at the correct SLOW speed.

But let's look at HIGH speed.

If M2 is a 12 pole machine, it's regular RPM would be 500(according to the formula: 120 x f _______ Np

Maybe it will run at 1000RPM by a suitable choice of capacitor?

If that is the case, we can see how the required 1000RPM would come from.

But, if that ran at 1000RPM it would transmit torque through the clutch to M1, the spindle of which could not be turned. You cannot try to turn M1 spindle from it;s spindle end, only from the rotor end because it's a geared motor.

I think you have to be quite inventive/clever to figure how this contraption works. :c)

Reply to
Richard

M2 is a very old 1940s design. I believe it it originally had a permanent magnet rotor and was used as a 12 pole. 24v stepper motor.

Your M2 looks like a rewound version probably with a 35% cobalt hysteresis armature. This would give it an approximately constant torque characteristic.

It's possible that it's not used as a motor but simply switched in to provide a constant braking torque to reduce the speed of M1.

jim

Reply to
pentagrid

According to your schematic, m1 doesn't have a starting cap. So, a synchronous motor, which isn't self starting and will run fine in either direction. Motors of this type were typically hand started by spinning the shaft (some old electric mains clocks), or used a mechanical spring/cam arrangement on the shaft to ensure auto staring in the correct direction. Have a look at this to see if it's gummed up, needs a clean and oil, whatever.

The second motor looks a like it is self starting via the 90 degree phase shift second winding and cap and may also be designed to start m1. You should be able to swap the wires on one or other of the two windings on M2 to get for/rev rotation.

Some of the older electro mechanical stuff could be pretty wierd, just to save a few parts :-)...

Regards,

Chris

Reply to
ChrisQ

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Another point is that your schematic shows the cap in parallel with the m2 winding, when it would normally be in series to the supply for this sort of arrangement...

Regards,

Chris

Reply to
ChrisQ

Chris has got it nearly right.

M1 looks like a typical synchronous motor as used in clocks and, probably has a mechanical "trip" that makes it start in a set direction. In older electric clocks, you often had a level out the back which you manually used to effect this. Later and better clocks had the "automatic" version. It will be a very low torque but geared so that the speed is reduced and torgue increased (just like a clock). I'd suggest from experienxce that this will provide a slow chart speed. BTW, the "trip" is like a pawl. If the motor starts correctly, then it simply "slips". If the motor, initilly, starts in the wrong direction, the "pawl" bites - stops the motor and the "hicough" makes it start again in the correct direction.

M2 is a typical synchronous (two phase) motor. Sorry Chris. The cap is in parallel. All you need is a phase shift ! If you have ever seen a Parvalux fractional HP motor as specified for the Quorn T&C Grinder, you'd have come acros higher powered versions. Connecting the capacitor across the other winding will reverse the motor direction. The synchronous speed will be 3000/6 ie 500 RPM if it has

12 poles. The value of the capacitor will NOT change the speed, just the torque to spme extent.

May I now guess that M2 has a through shaft ? If so, I suspect that the operation is as follows. M1 is a geared motor providing drive to the papaer (slow speed) by driving the set of gears visible on the RHS via the shaft of M2 which free-wheels. The clutch being engaged. When High-speed is selected by operation of the switch energsing M2, it drives the same set of gears in the same direction, but the clutch, now being "overdrived" slips. M1 continues to rotate. Just think about the mechaniical connexion between M1 and M2 and you will see what I mean.

Let me know if anything I've guessed was incorrect.

Reply to
houstonceng

M1 *is* a typical synchronous motor (Crouzet type 392). It is self starting and is marked as CL (clockwise). If I hold the clutch device at one end, torque is passed through the clutch and doing that makes the motor turn counter clockwise, however, it soon corrects itself and returns to clockwise direction. It is a geared motor 240V, 50 Hz and runs at 25 RPM.

M2 I believe is an hysterisis synchronous motor having a through-shaft. It appears that it has two windings that are referenced as "1" and "2", as well as "X" ad "Y".

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Above link we see that we can connect a capacitor *permanently* in series with one of the coils of an hysterisis synchronous motor - mains directly to the other coil.

In my circuit, the capacitor is connected in parallel with one of the coils which is different, but we must note that mains is now across both coils, unlike the usual connection.

As to operation of this contraption:

M1>---------------------------------SHAFT P Q

SLOW SPEED (Minutes)

Say M2 is not energised.

In order for M1 to properly drive the SHAFT it would need to rotate the SHAFT clockwise (looking into the SHAFT, i.e. looking to the RIGHT). M1 wants to rotate SHAFT counter clockwise because it's a clockwise motor. The clutch will transmit torque to M2 as M1's rotation is the right way to operate it. The gearing set to 1/2". I've measured the time it takes M1 to move the chart paper 1/2", which is 40 seconds. This speed is too fast, the SHAFT speed would need to be 16.67 RPM to move paper at 1/2" per minute.

So, these are the problems with M1 driving the SHAFT at SLOW speed.

HIGH SPEED (Seconds)

Multiplying 16.67 RPM by 60 would give the SHAFT speed for HIGH speed operation. That is 1000 RPM.

The synchronous speed of M2 would be 500 RPM if it has 12 poles which it has I'm sure. That is only half the required rotation of SHAFT in HIGH speed mode.

Let's assume M1 is still running.

The clutch will transmit torque when the one side is moving COUNTER CLOCKWISE" *relative to the other side*.

If M2 was to be rotating at 1000 RPM in a direction to make SHAFT rotate the correct way, point Q (see above) is the say the input side and it is moving 1025 RPM counter clockwise relative to point P.

Because torque is tranmitted from Q to P, (and also P to Q) M2 wants to overcome M1 and change it's speed from 25 RPM clockwise, to 1000 RPM counter clockise. In this situation M1 would win, making M2's spindle rotate at some speed lower than 25 RPM, and in the wrong direction for the SHAFT.

Really M1 should rotate at it's synchronous speed I think, not 16.67 RPM.

Yes indeed. If the SHAFT is meant to go the same way as M1 drives the SHAFT, and M2 drives the SHAFT the same way as M1, the thing works.

But, I've connected M1 direct to the SHAFT and the sprockets in the paper have the paper feeding the spool of paper, rather than drawing from it. That's the big spanner in the works.

Reply to
Richard

You are spot on!

It just came to me the error I was making. So simple.

The true picture is this:

SPINDLE M1>------------------------------X X ---SHAFT X

Bolting on M2 does not put it's axis concentric with SHAFT. The pinion on M2 spindle cogs with a pinion on SHAFT. And that will lead to a reversal of whatever way the M1 and M2's spindle turns. So, M1 will make sprokets move the correct way.

Also, this solves the speed problem. The pinion on the spindle is smaller than the one on SHAFT. SHAFT needs to revolve at 16.67 RPM, but the spindle needs to be running at a different speed, at 25 RPM, because cogs/pinions will reduce the RPM from 25 to 16.67 RPM.

What doea this do for the HIGH speedsituation? The spindle should rotate 60 times 25 RPM = 1500 RPM. SHAFT turns at 1000 RPM.

The only outstanding issue is how does M2 do 1500 RPM, if it's a 12 pole motor running at 50 Hz?

Reply to
Richard

Yes, the cap across the second winding would induce a phase shift in relation to the first and if the cap were arranged to resonate the winding, the phase shift would in fact be 90 degrees. Cap in series would still be valid though, so why is it done this way ?. Less wire in each winding as each then only has to handle half the line voltage ?...

Regards,

Chris

Reply to
ChrisQ

Although M2 has 12 stator pole pieces it is wound as a 4pole 1500 RPM machine. Numbering the slots 1 - 12 one pole winding occupies slots

1,2 5,6. The remaining windings following the same pattern fill all the slots

Jim

Reply to
pentagrid

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