I agree. Rotor inertia is necessary to drive the mechanical load in a
singlephase motor just as it is the third leg in an idler. I don't assert
that this is the dominant source of energy exchange, only that it is a
necessary contributor for at least part of each cycle.

Not that it matters in any practical way, but I'm still curious about how
much. I'm not curious enough (yet) to instrument it properly to measure
it.


wrote in message
...
On 11 Sep 2004 16:40:13 -0700, jim rozen
wrote:

*

In article , Don Foreman says...

This doesn't prove (or disprove) whether or not energy is withdrawn from
rotational energy during part of each cycle and replenished during a
different part of the same cycle. There is certainly countertorque
produced
when the rotor field is generating the third leg while excitation
voltage
(hence input excitation power) goes thru zero. I would agree that "the
....
energy built up during starting remains constant" as an average value,
but
that does not preclude significant alternating energy exhange during
each
cycle that nets to zero over each cycle under steady-state condx.

I can certainly believe that moment of inertia is already so high that
adding more produces no noticable improvement or difference. From a
practical standpoint, that's probably all that matters.

*

Once again I would suggest the thought experiment of having a
converter idler with a *zero* moment of inertia rotor. If
such an instrument would not function, because the rotor would
stop at the instant it is required to supply energy into the
third leg, then you are most likely correct - there is *some*
flywheel required.

But no more so than, say, the rotor in single phase motor supplys
to keep its rotor turning during the entire cycle.

Jim


It probably helps to look at limiting cases.

If there's zero mechanical inertia, because the
torque drops to zero twice per cycle, continuous rotation is
not possible.

If there is only JUST enough inertia to enable
rotation to continue through the low torque parts of the
supply waveform, the rotational velocity is not constant and
drops towards zero during this low torque period.

In the period that it remains below its long term
average speed, as the rotor slows, it tries to take more
power from the supply. In the period that it is above the
long term average it returns the excess power to the supply.

Long term (ignoring pesky second order effects) its
input/output power balance is the same as as a motor with
very large inertia.

A second order effect that cannot be ignored is
the effect on the back EMF waveform. Because of the velocity
variation this is no longer sinusoidal and this results in a
large third harmonic content in the current drawn from the
supply. For the same reason, if this very low inertia motor
is used as a rotary converter, the phantom phase voltage
waveform will be distorted and contain a large third
harmonic component.

Very fortunately, for the sort of motors that we we
use, rotor mechanical inertias are so large that the effects
fade into insignificance. The torque fluctuations are so
well averaged by the rotor inertia that the only effect we
observe is the slight vibration of the frame of a single
phase driven motor as it tries to accelerate and decelerate
the rotor . The effect on phantom phase waveform is equally
small and masked by the normal harmonic content of a
commercial supply.

Jim

On Sat, 11 Sep 2004 16:11:10 -0400, Gary Coffman
wrote:
*
On Sat, 11 Sep 2004 15:58:26 +0000 (UTC), wrote:
The output pattern of these three phase winding
can be directly measured and show that the stator generates
a magnetic field pattern in a squirrel cage rotor that
rotates at slip frequency in the same direction as the
mechanical rotation.

Because, in our real motor, this rotation of the
magnetic field orientation is additional to and adds to the
mechanical rotation it exactly cancels the slip speed and
returns the phantom phase output to the correct frequency.

Both the mechanical rotation and the magnetic
correction are resisting the same drag torque from ouput
loading so the energy contributions are in proportion to
their speed - about 95% from the motor plus generator
mechanical rotation and 5% for slowly dragging the stored
field round the rotor.


*


This is where I have a problem. In a steady state system,
the output energy equals the input energy (minus system
losses). The mechanically stored rotational energy built up
during motor starting remains constant (if it didn't, the rotor
would slow to a stop). So the mechanical rotation contributes
*zero* to the output energy. It is only the energy coupled via
the slipping field which is transferred through the system.
The mechanical rotation only establishes *phasing*. It isn't
the source of energy for the output.

This may appear to be a fine semantic distinction, but it
explains why adding flywheel mass to a rotary converter
doesn't help.

Gary


I think this is basically semantics and the usual
problems of trying to keep posts reasonably short.

It's really the result of trying to use single step
and two step models to describe the same process. I prefer
the two step model because, by conceptually separating the
portion of the input current that produces torque, from the
portion that rotates the orientation of the magnetic field
in the rotor, it produces an easily assimilated picture of
what's happening. Of course, as you say, there's only one
real current and this produces the slipping field that you
describe.

The two step model does NOT require the rotational
energy stored in the rotor INERTIA to make any long term
contribution to the output energy but it makes clear that
mechanical ROTATION of the rotor is an essential part of the
phantom phase generation. If the rotor stops there is no
energy transfer.

The same two step model covers operation both with
the input power power present or removed and it neatly and
exactly explains the surprisingly small drop in phantom
phase output power and frequency for the first few revs
after the power input completely removed.

The single step model can also explain the same process but
it is much more difficult. Once the supply is removed the
slipping field transfer is no longer operative and it is
necessary to now take into account the effect of the still
remaining rotor magnetic field that is the core of the two
step model.

It's clear that I overdid trying to keep the post
short and simple. The following extended penultimate
paragraph perhaps better explains what I was trying to say.



Both the mechanical rotation RESULTING FROM THE STATOR
INPUT CURRENT PROVIDING DRIVING TORQUE TO THE ROTOR and the
magnetic correction are resisting the same drag torque from
ouput loading. The energy contributions are in proportion to
their speed - about 95% from the motor plus generator
mechanical rotation and 5% for slowly dragging the stored
field round the rotor.

Jim

On Sun, 12 Sep 2004 09:34:59 +0000 (UTC), wrote:
On 11 Sep 2004 16:40:13 -0700, jim rozen
wrote:
Once again I would suggest the thought experiment of having a
converter idler with a *zero* moment of inertia rotor. If
such an instrument would not function, because the rotor would
stop at the instant it is required to supply energy into the
third leg, then you are most likely correct - there is *some*
flywheel required.

But no more so than, say, the rotor in single phase motor supplys
to keep its rotor turning during the entire cycle.

Jim


It probably helps to look at limiting cases.

If there's zero mechanical inertia, because the
torque drops to zero twice per cycle, continuous rotation is
not possible.

If there is only JUST enough inertia to enable
rotation to continue through the low torque parts of the
supply waveform, the rotational velocity is not constant and
drops towards zero during this low torque period.

In the period that it remains below its long term
average speed, as the rotor slows, it tries to take more
power from the supply. In the period that it is above the
long term average it returns the excess power to the supply.

In other words, the slip changes.

Long term (ignoring pesky second order effects) its
input/output power balance is the same as as a motor with
very large inertia.

Yes.

Gary

All this talk got me to reading my audels electric motor book.
They state that a synchronous motor is almost identical to
an alternator. Since I have a 3 PH alternator, I was wondering
how hard it would be to make it run as a sync motor and once
it is running to use it as a rotary phase converter. This would
be more efficient than driving it with a single phase motor.

My limited research seems to indicate that I need to get it
very close to sync speed with a pony motor and then apply
power to two phases and DC to the field. Then it should
motor just like a motor and hopefully the third leg will be
generating. Yes or NO?

chuck

On 13 Sep 2004 18:19:17 GMT,
(Charles A. Sherwood) wrote:

All this talk got me to reading my audels electric motor book.
They state that a synchronous motor is almost identical to
an alternator. Since I have a 3 PH alternator, I was wondering
how hard it would be to make it run as a sync motor and once
it is running to use it as a rotary phase converter. This would
be more efficient than driving it with a single phase motor.

My limited research seems to indicate that I need to get it
very close to sync speed with a pony motor and then apply
power to two phases and DC to the field. Then it should
motor just like a motor and hopefully the third leg will be
generating. Yes or NO?

chuck

This should be a very interesting experiment and inject some
hard facts into theoretical speculations.

The two step model suggests that it will behave exactly
as you suggest - once operating as a single phase driven
synchronous motor, the rotating DC excited rotor will
generate the phantom phase but a bit low on volts because of
leakage inductance and resistive losses.

One thing you will need to watch is the level of the DC
excitation. Because the rotor is synchronous it can only run
at a single fixed speed. The back EMF will be directly
proportional to field strength and operation should be with
the back EMF a bit less (a "bit" to allow for unknown
losses) than supply volts minus winding voltage drop.

Too much field causes the motor to draw excess
leading current from the supply (it's behaving as a
capacitor) too little causes excess lagging current
resulting in inductive behaviour.

These comments are based on the behaviour of three
phase excited synchronous motors. I'd expect the single
phase excited machine to be at least roughly similar.

Do let us know how you get on.

Jim

In article ,
wrote:

[...]

This should be a very interesting experiment and inject some
hard facts into theoretical speculations.

The two step model suggests that it will behave exactly
as you suggest - once operating as a single phase driven
synchronous motor, the rotating DC excited rotor will
generate the phantom phase but a bit low on volts because of
leakage inductance and resistive losses.

One thing you will need to watch is the level of the DC
excitation. Because the rotor is synchronous it can only run
at a single fixed speed. The back EMF will be directly
proportional to field strength and operation should be with
the back EMF a bit less (a "bit" to allow for unknown
losses) than supply volts minus winding voltage drop.

Too much field causes the motor to draw excess
leading current from the supply (it's behaving as a
capacitor) too little causes excess lagging current
resulting in inductive behaviour.

These comments are based on the behaviour of three
phase excited synchronous motors. I'd expect the single
phase excited machine to be at least roughly similar.

Do let us know how you get on.

Jim

I know this is a divergent train of thought, but could one short the
slip rings' leads together and make it a strict induction motor? I
don't suppose it would be efficient at all.

--
B.B. --I am not a goat! thegoat4 at airmail.net

On 13 Sep 2004 18:19:17 GMT, (Charles A. Sherwood) wrote:
All this talk got me to reading my audels electric motor book.
They state that a synchronous motor is almost identical to
an alternator. Since I have a 3 PH alternator, I was wondering
how hard it would be to make it run as a sync motor and once
it is running to use it as a rotary phase converter. This would
be more efficient than driving it with a single phase motor.

My limited research seems to indicate that I need to get it
very close to sync speed with a pony motor and then apply
power to two phases and DC to the field. Then it should
motor just like a motor and hopefully the third leg will be
generating. Yes or NO?

Should work, but there won't be any automatic voltage regulation
of the phantom leg with varying load (slip does that for you in an
ordinary squirrel cage RPC). You'll need to dynamically vary the
field current to regulate the 3rd leg voltage with changing
loads. Alternator regulators normally do this, but configuring one
to only look at the 3rd leg voltage might cause a bit of head
scratching.

Gary

This sounds neat. Why would it be hard to configure a regulator to look at
the third leg?

"Gary Coffman" wrote in message
...


Should work, but there won't be any automatic voltage regulation
of the phantom leg with varying load (slip does that for you in an
ordinary squirrel cage RPC). You'll need to dynamically vary the
field current to regulate the 3rd leg voltage with changing
loads. Alternator regulators normally do this, but configuring one
to only look at the 3rd leg voltage might cause a bit of head
scratching.

Gary

On Mon, 13 Sep 2004 17:27:35 -0500, "B.B."
u wrote:



In article ,
wrote:

[...]

This should be a very interesting experiment and inject some
hard facts into theoretical speculations.

The two step model suggests that it will behave exactly
as you suggest - once operating as a single phase driven
synchronous motor, the rotating DC excited rotor will
generate the phantom phase but a bit low on volts because of
leakage inductance and resistive losses.

One thing you will need to watch is the level of the DC
excitation. Because the rotor is synchronous it can only run
at a single fixed speed. The back EMF will be directly
proportional to field strength and operation should be with
the back EMF a bit less (a "bit" to allow for unknown
losses) than supply volts minus winding voltage drop.

Too much field causes the motor to draw excess
leading current from the supply (it's behaving as a
capacitor) too little causes excess lagging current
resulting in inductive behaviour.

These comments are based on the behaviour of three
phase excited synchronous motors. I'd expect the single
phase excited machine to be at least roughly similar.

Do let us know how you get on.

Jim





I know this is a divergent train of thought, but could one short the
slip rings' leads together and make it a strict induction motor? I
don't suppose it would be efficient at all.





It depends on the shape of the pole pieces. It might
be possible to make it work as a salient pole type
synchronous motor (pretty similar in principle to some early
stepper motor types).

It's unlikely to work as as a straightforward induction
motor because this needs the circularly symmetric pattern
formed by squirrel cage rotor bars or a properly distributed
three phase winding.

Jim

I think this is intriguing, might be worth a bit of head scratching. It's
intriguing because it offers the possibility of a "self-tuning" RPC that
adapts to varying loads. Electronics could be quite simple because they
need not produce significant AC power (as in a VFD) but only control DC
excitation current.

Assuming that a neutral point is not available, a neutral-equivalent voltage
reference could be synthesized with opamps. It would be half the
line-to-line excitation plus a quadrature component produced by an RC
phaseshift circuit. Perhaps the third-leg phase voltage relative to this
quasi-neutral could then be compared to the other phases relative to the
same point, and excitation then self-adjusted to minimize the difference
between generated third-leg magnitude and the other phase magnitudes
relative to the quasi-neutral. Generated third-leg would be whatever it
turns out to be, since we have only one controlled variable -- excitation.
However, if the magnitude is right then I think the phase would be pretty
close to right due to the geometry of the machine.

Such a controller could be accomplished digitally with a microcomputer of
course, but I'd find it easier to see and understand what's going on in
a controller realized with a few opamps.

Comments? I could be up for helping to design and even build such a gadget
just to see if and how it works.

"Gary Coffman" wrote in message
...


Should work, but there won't be any automatic voltage regulation
of the phantom leg with varying load (slip does that for you in an
ordinary squirrel cage RPC). You'll need to dynamically vary the
field current to regulate the 3rd leg voltage with changing
loads. Alternator regulators normally do this, but configuring one
to only look at the 3rd leg voltage might cause a bit of head
scratching.

Gary

On Tue, 14 Sep 2004 09:11:24 +0000 (UTC), "Don Foreman"
wrote:

I think this is intriguing, might be worth a bit of head scratching. It's
intriguing because it offers the possibility of a "self-tuning" RPC that
adapts to varying loads. Electronics could be quite simple because they
need not produce significant AC power (as in a VFD) but only control DC
excitation current.

Assuming that a neutral point is not available, a neutral-equivalent voltage
reference could be synthesized with opamps. It would be half the
line-to-line excitation plus a quadrature component produced by an RC
phaseshift circuit. Perhaps the third-leg phase voltage relative to this
quasi-neutral could then be compared to the other phases relative to the
same point, and excitation then self-adjusted to minimize the difference
between generated third-leg magnitude and the other phase magnitudes
relative to the quasi-neutral. Generated third-leg would be whatever it
turns out to be, since we have only one controlled variable -- excitation.
However, if the magnitude is right then I think the phase would be pretty
close to right due to the geometry of the machine.

Such a controller could be accomplished digitally with a microcomputer of
course, but I'd find it easier to see and understand what's going on in
a controller realized with a few opamps.

Comments? I could be up for helping to design and even build such a gadget
just to see if and how it works.


Should be an interesting project.

A possible alternative method of synthesising the signal is
a couple of low power transformers in Scott connection. The
first is a centre tapped auto across the supply. The second
is of any convenient ratio with primary connected from
centre tap to phantom phase.

Transformers are a nuisance but they have the big
advantage that they fully isolate the control circuitry from
the high power bits.

If isolation isn't important, an opamp looking at
the difference between a three equal resistor artificial
neutral and the phantom phase is yet another way.

The really interesting measurement will be to see
what happens to the phantom phase angle when the regulator
corrects the voltage drop resulting from a heavy load. If
the model really describes the way it works we should see
the effect of a small shift in the angular relation between
the rotating field magnet and the rotating field component
of the supply.

There's a good chance that it will be a really
useful improvement on the basic rotary converter but it's
pity that 3 phase alternators are a lot rarer than motors.

Jim

Isolation transformers would definitely be a good thing.

wrote in message
news
On Tue, 14 Sep 2004 09:11:24 +0000 (UTC), "Don Foreman"
wrote:

I think this is intriguing, might be worth a bit of head scratching.
It's
intriguing because it offers the possibility of a "self-tuning" RPC that
adapts to varying loads. Electronics could be quite simple because they
need not produce significant AC power (as in a VFD) but only control DC
excitation current.

Assuming that a neutral point is not available, a neutral-equivalent
voltage
reference could be synthesized with opamps. It would be half the
line-to-line excitation plus a quadrature component produced by an RC
phaseshift circuit. Perhaps the third-leg phase voltage relative to
this
quasi-neutral could then be compared to the other phases relative to the
same point, and excitation then self-adjusted to minimize the difference
between generated third-leg magnitude and the other phase magnitudes
relative to the quasi-neutral. Generated third-leg would be whatever
it
turns out to be, since we have only one controlled variable --
excitation.
However, if the magnitude is right then I think the phase would be
pretty
close to right due to the geometry of the machine.

Such a controller could be accomplished digitally with a microcomputer
of
course, but I'd find it easier to see and understand what's going on
in
a controller realized with a few opamps.

Comments? I could be up for helping to design and even build such a
gadget
just to see if and how it works.


Should be an interesting project.

A possible alternative method of synthesising the signal is
a couple of low power transformers in Scott connection. The
first is a centre tapped auto across the supply. The second
is of any convenient ratio with primary connected from
centre tap to phantom phase.

Transformers are a nuisance but they have the big
advantage that they fully isolate the control circuitry from
the high power bits.

If isolation isn't important, an opamp looking at
the difference between a three equal resistor artificial
neutral and the phantom phase is yet another way.

The really interesting measurement will be to see
what happens to the phantom phase angle when the regulator
corrects the voltage drop resulting from a heavy load. If
the model really describes the way it works we should see
the effect of a small shift in the angular relation between
the rotating field magnet and the rotating field component
of the supply.

There's a good chance that it will be a really
useful improvement on the basic rotary converter but it's
pity that 3 phase alternators are a lot rarer than motors.

Jim

Assuming that a neutral point is not available, a neutral-equivalent voltage


The alternator is a 12 wire, so it can be connected as delta or wye.
If connected as wye, a netural point is available. IF connected as
delta, the voltage from either leg to the wild leg indicates the
wild leg voltage. Seems straight forward?

Also keep in mind it is a brushless alternator which means there
are really two alternators wrapped up in one. The excitor has a field
that is stationary. The rotor on the excitor is a 3 phase alternator
that produces AC (at 120Hz because it has 6 poles) which is rectified
by diodes on the rotor. The DC from the diodes powers the main
rotating field. It only takes 30 volts at about 1 amp on the excitor
field to drive the alternator to 15kw.

When used as an alternator driven by a engine, the electronic voltage
regulator is powered by 2 legs and drives the excitor field. It is
really only regulating line 1 to line 2 and line 3 gives what it
gives. Seems like the sense point could be moved to line 1 to wild
leg.

Anyway my point here is there is a huge time constant involved
with regulating the output going through all inductance.

On 14 Sep 2004 22:20:14 GMT,
(Charles A. Sherwood) wrote:

Assuming that a neutral point is not available, a neutral-equivalent voltage


The alternator is a 12 wire, so it can be connected as delta or wye.
If connected as wye, a netural point is available. IF connected as
delta, the voltage from either leg to the wild leg indicates the
wild leg voltage. Seems straight forward?

Also keep in mind it is a brushless alternator which means there
are really two alternators wrapped up in one. The excitor has a field
that is stationary. The rotor on the excitor is a 3 phase alternator
that produces AC (at 120Hz because it has 6 poles) which is rectified
by diodes on the rotor. The DC from the diodes powers the main
rotating field. It only takes 30 volts at about 1 amp on the excitor
field to drive the alternator to 15kw.

When used as an alternator driven by a engine, the electronic voltage
regulator is powered by 2 legs and drives the excitor field. It is
really only regulating line 1 to line 2 and line 3 gives what it
gives. Seems like the sense point could be moved to line 1 to wild
leg.

Anyway my point here is there is a huge time constant involved
with regulating the output going through all inductance.



Sounds logical and well worth a try.

The lags involved in the time constants of a brushless
alternator set up make it a bit more difficult to stabilise
a regulator loop. However, if we've guessed right, the loop
gain and lags of the revised regulator loop should be
sufficiently similar to the previous connection for it
remain stable.

Moving the sense connection to wild leg has the major
advantage of minimum change to the regulator loop.
Theoretically it would be better to sense from a real or
synthetic neutral but I don't think it's different enough to
matter.

What is difficult to predict is the transient
behaviour as it first tries to lock into synchronous
rotation. Probably better to check it out first with fixed
excitation before allowing the regulator to take charge.

Jim

On Tue, 14 Sep 2004 08:04:08 +0000 (UTC), "Don Foreman" wrote:
This sounds neat. Why would it be hard to configure a regulator to look at
the third leg?

Because a standard 3 ph alternator regulator doesn't work that way.
You don't want it trying to regulate the two legs supplied by the utility,
and it may be non-trivial to convince a standard regulator not to try to
do so. Of courseyou can scratch build a custom regulator. Don Foreman
had some good ideas on that.

Gary

Probably would be a good idea to characterize it with a few experiments, as:

give excitation a step change and observe the time response of the generated
voltage.

Think the dynamics would change with load?


wrote in message
...
On 14 Sep 2004 22:20:14 GMT,
(Charles A. Sherwood) wrote:

Assuming that a neutral point is not available, a neutral-equivalent
voltage


The alternator is a 12 wire, so it can be connected as delta or wye.
If connected as wye, a netural point is available. IF connected as
delta, the voltage from either leg to the wild leg indicates the
wild leg voltage. Seems straight forward?

Also keep in mind it is a brushless alternator which means there
are really two alternators wrapped up in one. The excitor has a field
that is stationary. The rotor on the excitor is a 3 phase alternator
that produces AC (at 120Hz because it has 6 poles) which is rectified
by diodes on the rotor. The DC from the diodes powers the main
rotating field. It only takes 30 volts at about 1 amp on the excitor
field to drive the alternator to 15kw.

When used as an alternator driven by a engine, the electronic voltage
regulator is powered by 2 legs and drives the excitor field. It is
really only regulating line 1 to line 2 and line 3 gives what it
gives. Seems like the sense point could be moved to line 1 to wild
leg.

Anyway my point here is there is a huge time constant involved
with regulating the output going through all inductance.



Sounds logical and well worth a try.

The lags involved in the time constants of a brushless
alternator set up make it a bit more difficult to stabilise
a regulator loop. However, if we've guessed right, the loop
gain and lags of the revised regulator loop should be
sufficiently similar to the previous connection for it
remain stable.

Moving the sense connection to wild leg has the major
advantage of minimum change to the regulator loop.
Theoretically it would be better to sense from a real or
synthetic neutral but I don't think it's different enough to
matter.

What is difficult to predict is the transient
behaviour as it first tries to lock into synchronous
rotation. Probably better to check it out first with fixed
excitation before allowing the regulator to take charge.

Jim

On Wed, 15 Sep 2004 09:40:37 +0000 (UTC), "Don Foreman"
wrote:

Probably would be a good idea to characterize it with a few experiments, as:

give excitation a step change and observe the time response of the generated
voltage.

Think the dynamics would change with load?



Would be a good move - what would be really nice is to
look at the short term changes with a storage scope.

Not sure how the dynamics will change with load - hopefully
not a lot. There are so many second order effects!

One interesting one is the effect of hysteresis in the
soft iron field circuit of the exciter alternator - sort of
minor electronic backlash within the servo loop. However my
present feeling is that, whoever designed the existing
regulator loop will have already solved most of these
problems for us.

Jim

wrote in message
...

Probably would be a good idea to characterize it with a few experiments,
as:

give excitation a step change and observe the time response of the
generated
voltage.

Think the dynamics would change with load?



Would be a good move - what would be really nice is to
look at the short term changes with a storage scope.

Yup. That or a PC-based data-acq setup. I have a scope and a LabJack DAQ
but no synchronous motor. DATAQ offers a simple DAQ for $24.95.
http://www.dataq.com/products/startkit/di194rs.htm
It's kinda slow at 240 samples/sec. That might be fast enough, given the
large inductances and rotational inertias involved.

Not sure how the dynamics will change with load - hopefully
not a lot. There are so many second order effects!

One interesting one is the effect of hysteresis in the
soft iron field circuit of the exciter alternator - sort of
minor electronic backlash within the servo loop. However my
present feeling is that, whoever designed the existing
regulator loop will have already solved most of these
problems for us.

Solved them, or just ignored them. Many control systems just do the best
they can with the system at hand, warts 'n all. Consider, for example, an
automotive cruise control. They certainly aren't quick and they're not
always precise, but they usually work acceptably well.

On Wed, 15 Sep 2004 15:38:18 +0000 (UTC), "Don Foreman"
wrote:


wrote in message
.. .


snip
Would be a good move - what would be really nice is to
look at the short term changes with a storage scope.




Yup. That or a PC-based data-acq setup. I have a scope and a LabJack DAQ
but no synchronous motor. DATAQ offers a simple DAQ for $24.95.
http://www.dataq.com/products/startkit/di194rs.htm
It's kinda slow at 240 samples/sec. That might be fast enough, given the
large inductances and rotational inertias involved.

snip

The trouble is that suitable machines are pretty thin on
the ground.I've got a pretty ancient storage scope but it's
the wrong side of the atlantic!

I don't know if advice from our photographic
friends would help. I believe some of the better digital
cameras have a 5 second maximum shutter time - would one of
these do the necessary with a standard scope?

Jim

I've done that, using an Olympus C2500L camera. It works OK if the scope
screen is hooded to cut out reflections and glare from the environment.

However, for a 5 second period, I think the $24.95 DATAQ might work better.

wrote in message
...
On Wed, 15 Sep 2004 15:38:18 +0000 (UTC), "Don Foreman"

Yup. That or a PC-based data-acq setup. I have a scope and a LabJack
DAQ
but no synchronous motor. DATAQ offers a simple DAQ for $24.95.
http://www.dataq.com/products/startkit/di194rs.htm
It's kinda slow at 240 samples/sec. That might be fast enough, given
the
large inductances and rotational inertias involved.

snip

The trouble is that suitable machines are pretty thin on
the ground.I've got a pretty ancient storage scope but it's
the wrong side of the atlantic!

I don't know if advice from our photographic
friends would help. I believe some of the better digital
cameras have a 5 second maximum shutter time - would one of
these do the necessary with a standard scope?

Jim

Maybe I've got a crazy idea or have I had too many, but let's say that I buy a
Bridgeport or some old lathe, boring mill, or surface grinder that has a 3ph
motor. The idea is to get this machine running on my 1ph service as cheaply as
possible. Why not buy some big clunky 1 or 3 ph motor from the boneyard, use
the apropiate circuit to make it run and couple it up to an automotive
alternator with the diodes bypassed to get 3ph AC.Most automotive alternators
will put out 100a at 12v but 10a at 120v will still handle a decent sized
motor. I'd think that it would be a good idea to have the motor running the
alternator to have the same speed as the motor being run from it. This would
eliminate the need for pony motors although the alternator would require a 12v
DC field power supply. This would also be an advantage because low voltage
wiring to a switch could allow you to "freewheel" your alternator when not
running your machine.
Engineman1

On 19 Sep 2004 01:15:47 GMT, (Engineman1) wrote:
Maybe I've got a crazy idea or have I had too many, but let's say that I buy a
Bridgeport or some old lathe, boring mill, or surface grinder that has a 3ph
motor. The idea is to get this machine running on my 1ph service as cheaply as
possible. Why not buy some big clunky 1 or 3 ph motor from the boneyard, use
the apropiate circuit to make it run and couple it up to an automotive
alternator with the diodes bypassed to get 3ph AC.Most automotive alternators
will put out 100a at 12v but 10a at 120v will still handle a decent sized
motor. I'd think that it would be a good idea to have the motor running the
alternator to have the same speed as the motor being run from it. This would
eliminate the need for pony motors although the alternator would require a 12v
DC field power supply. This would also be an advantage because low voltage
wiring to a switch could allow you to "freewheel" your alternator when not
running your machine.
Engineman1

There are a few problems with that approach. First, an auto alternator
typically can't produce more than about 130 volts, and that only at high
RPM. Since nearly all machine tool 3 ph motors expect 220 volts, you
won't have enough voltage to operate them properly.

Second, you can't assume that a 12 volt 100 amp alternator can produce
10 amps at 120 volts. Output voltage is proportional to rotor current, and
typical auto alternators are only rated for 3 amps of rotor current. At 12
volts, the typical alternator requires 1 amp of rotor current to produce
12 volts at rated output. If that scales linearly with voltage, you'd need
10 amps of rotor current to make 10 amps output at 120 volts. That's
3 times the rated rotor current.

Finally, to get 60 Hz 3 ph, the alternator has to turn at a specific RPM.
For a 2 pole alternator, that's 3600 RPM, well below the RPM where the
alternator makes maximum output. (In a car, the pulleys are sized so
that the alternator turns faster than engine RPM to get it up to the speed
where it produces maximum output.) Output frequency doesn't matter in
the auto application because the output is rectified to DC anyway. But
it does matter if you're going to drive a 3 ph AC motor with it.

Gary