On 10 Mar, 11:24, Thomas Prufer
wrote:
On Mon, 10 Mar 2008 03:35:00 -0700 (PDT), The Real Doctor

wrote:
However, I'd like to know how you measured the power in this case.
Measurements on very jagged waveforms (ie very high bandwidth) are
notoriously tricky to do, as the peddlers of many perpertual motion
machines have strenuously resisted finding out.

One thing to measure accurately for purposes of knowledge, scientific
advancement, or the satisfaction of curiosity, another to devise a waveform
which will allow power to be transferred without a given power meter indicating
accordingly.

It's possible that the perpetual motion con artists go to that degree
of trouble, but I don;t think they are bright enough in general. I
suspect that they may well try several power meters before deciding
which to demonstrate to their marks.

ISTR some very modern rollercoaster -- using linear motors and very high surge
currents -- registered surprisingly little electricity use. Eventually they
adapted the meters accordingly.

The electricity companies get very cross if one strays too far from a
power factor of 1.0 on a domestic meter. My recollection is that they
used to allow up to 0.9 lagging and no lead at all, but I am open to
correction.

Ian

On Mon, 10 Mar 2008 12:12:13 +0000, Derek Geldard
wrote:

On Mon, 10 Mar 2008 10:46:02 +0000, Chris Charles
wrote:



Sort of..sort of not..

It chops the waveform. So the current is definitely out of phase with
the voltage..bein zero some of the time,and in phase the rest.

I'd hesitate to say what that means in terms of power factor though.

That is a term that is applied to liner combinations of L, C and R,
which a dimmer most certainly is not.



Why is the current out ofm phase with the voltage?

The filament is overwhelmingly resitive so the current and votage must
be in phase.



Even if the current is not switched on until it's 90 degrees out of
phase ?

Current can't flow until there is a circuit - switched on - and a
voltage is applied. If the circuit is resistive then the current is in
phase with the votage. It can't be any other way for a resistive load.

No voltage = no current however the voltade is applied.

That's so inaccurate as to be no use whatsoever.

??? If the load is resistive - as the filament is - then no voltage =
no current. see basic ohms law. I=E/R for a resistive circuit.

On Mon, 10 Mar 2008 08:05:46 -0700 (PDT), The Real Doctor
wrote:



ISTR some very modern rollercoaster -- using linear motors and very high surge
currents -- registered surprisingly little electricity use. Eventually they
adapted the meters accordingly.

The electricity companies get very cross if one strays too far from a
power factor of 1.0 on a domestic meter. My recollection is that they
used to allow up to 0.9 lagging and no lead at all, but I am open to
correction.


Ho - Ho. (weedy pun).

DG

On Mon, 10 Mar 2008 08:01:28 -0700 (PDT), The Real Doctor
wrote:

On 10 Mar, 12:12, Derek Geldard wrote:
On Mon, 10 Mar 2008 10:46:02 +0000, Chris Charles
wrote:

The filament is overwhelmingly resitive so the current and votage must
be in phase.

Even if the current is not switched on until it's 90 degrees out of
phase ?

I think you are getting a little confused here. Sketch the voltage
waveform produced by a dimmer. Now sketch the current waveform you
would expect if that voltage waveform went into a purely resistive
load. Hint: I(t) = V(t)/R.


Well even that's not right, dI/dt cannot be infinite and I is at zero
to start with, and also every time the "perfect switch" closes every
half cycle.

That aside, if you were to carry out a fourier series analysis on the
resultant I waveform you would find it has a strong component at the
50 Hz. fundamental (plus plenty of harmonics) and a constant delay
term related to the phase angle at which the "perfect switch" closes.

It is this phase angle which governs the power transmitted by the
dimmer, and also causes the current to lag the voltage.

I do restrict the discussion to cheap, simple phase angle devices such
as lamp dimmers and drill speed contollers. If there are "true
sinewave" dimmers (as there are "true sinewave" inverters.) I've not
come across any, and probably couldn't afford them if I did.

DG

On Mar 10, 5:21 pm, Derek Geldard wrote:
On Mon, 10 Mar 2008 08:01:28 -0700 (PDT), The Real Doctor



wrote:
On 10 Mar, 12:12, Derek Geldard wrote:
On Mon, 10 Mar 2008 10:46:02 +0000, Chris Charles
wrote:

The filament is overwhelmingly resitive so the current and votage must
be in phase.

Even if the current is not switched on until it's 90 degrees out of
phase ?

I think you are getting a little confused here. Sketch the voltage
waveform produced by a dimmer. Now sketch the current waveform you
would expect if that voltage waveform went into a purely resistive
load. Hint: I(t) = V(t)/R.

Well even that's not right, dI/dt cannot be infinite and I is at zero
to start with, and also every time the "perfect switch" closes every
half cycle.

That aside, if you were to carry out a fourier series analysis on the
resultant I waveform you would find it has a strong component at the
50 Hz. fundamental (plus plenty of harmonics) and a constant delay
term related to the phase angle at which the "perfect switch" closes.

It is this phase angle which governs the power transmitted by the
dimmer, and also causes the current to lag the voltage.

I do restrict the discussion to cheap, simple phase angle devices such
as lamp dimmers and drill speed contollers. If there are "true
sinewave" dimmers (as there are "true sinewave" inverters.) I've not
come across any, and probably couldn't afford them if I did.

DG

Sine wave dimmers are available from a few manufacturers, heres one:

http://www.silentdimming.com/index.asp

If you have to ask the price.....

Adam

That aside, if you were to carry out a fourier series analysis on the
resultant I waveform you would find it has a strong component at the
50 Hz. fundamental (plus plenty of harmonics) and a constant delay
term related to the phase angle at which the "perfect switch" closes.

It is this phase angle which governs the power transmitted by the
dimmer, and also causes the current to lag the voltage.

I do restrict the discussion to cheap, simple phase angle devices such
as lamp dimmers and drill speed contollers. If there are "true
sinewave" dimmers (as there are "true sinewave" inverters.) I've not
come across any, and probably couldn't afford them if I did.

DG

Nonsense


You cannot have a power factor with a resistive load. By definition.

The current and voltage are always in phase.

On Mon, 10 Mar 2008 19:36:37 +0000, Chris Charles
wrote:


That aside, if you were to carry out a fourier series analysis on the
resultant I waveform you would find it has a strong component at the
50 Hz. fundamental (plus plenty of harmonics) and a constant delay
term related to the phase angle at which the "perfect switch" closes.

It is this phase angle which governs the power transmitted by the
dimmer, and also causes the current to lag the voltage.

I do restrict the discussion to cheap, simple phase angle devices such
as lamp dimmers and drill speed contollers. If there are "true
sinewave" dimmers (as there are "true sinewave" inverters.) I've not
come across any, and probably couldn't afford them if I did.

DG

Nonsense


You cannot have a power factor with a resistive load. By definition.


Your definition does not encompass active components.

The current and voltage are always in phase.

Clearly, nay obviously, they are not.

DG

On Mon, 10 Mar 2008 19:48:38 +0000, Derek Geldard

I do restrict the discussion to cheap, simple phase angle devices such
as lamp dimmers and drill speed contollers. If there are "true
sinewave" dimmers (as there are "true sinewave" inverters.) I've not
come across any, and probably couldn't afford them if I did.

DG

Nonsense


You cannot have a power factor with a resistive load. By definition.


Your definition does not encompass active components.

The current and voltage are always in phase.

Clearly, nay obviously, they are not.


Ok, if that is so what is producing the phase shift.

On 2008-03-10 10:36:48 +0000, The Real Doctor said:

On 6 Mar, 13:45, David Hansen wrote:

Nice try

Yes! We used to count David's "Nice try" postings over on uk.railway,
but gave up when he got to several hundred, Does he do "Excellent.
Personal abuse." here as well?

Bless him.

Ian

Oh, he's not a trainspotter as well, is he?

On Mon, 10 Mar 2008 20:41:51 +0000, Chris Charles
wrote:

On Mon, 10 Mar 2008 19:48:38 +0000, Derek Geldard

I do restrict the discussion to cheap, simple phase angle devices such
as lamp dimmers and drill speed contollers. If there are "true
sinewave" dimmers (as there are "true sinewave" inverters.) I've not
come across any, and probably couldn't afford them if I did.

DG

Nonsense


You cannot have a power factor with a resistive load. By definition.


Your definition does not encompass active components.

The current and voltage are always in phase.

Clearly, nay obviously, they are not.


Ok, if that is so what is producing the phase shift.

It's implicit in there being a delay between the voltage impressed and
the current drawn.

DG

On 10 Mar, 17:21, Derek Geldard wrote:
On Mon, 10 Mar 2008 08:01:28 -0700 (PDT), The Real Doctor

I think you are getting a little confused here. Sketch the voltage
waveform produced by a dimmer. Now sketch the current waveform you
would expect if that voltage waveform went into a purely resistive
load. Hint: I(t) = V(t)/R.

Well even that's not right, dI/dt cannot be infinite and I is at zero
to start with, and also every time the "perfect switch" closes every
half cycle.

Tiny, tiny second order effect.

That aside, if you were to carry out a fourier series analysis on the
resultant I waveform you would find it has a strong component at the
50 Hz. fundamental (plus plenty of harmonics) and a constant delay
term related to the phase angle at which the "perfect switch" closes.

I am not quite sure what you are getting at here (and not because I
don't understand Fourier analysis, by the way - I've written a thesis
on it). How do you think the current and voltage waveforms in a
resistive load are related?

It is this phase angle which governs the power transmitted by the
dimmer, and also causes the current to lag the voltage.

With all dues respect ... cobblers.

Ian

On 10 Mar, 22:09, Derek Geldard wrote:

It's implicit in there being a delay between the voltage impressed and
the current drawn.

What voltage?

Ian

Derek Geldard wrote:
On Mon, 10 Mar 2008 08:01:28 -0700 (PDT), The Real Doctor
wrote:

On 10 Mar, 12:12, Derek Geldard wrote:
On Mon, 10 Mar 2008 10:46:02 +0000, Chris Charles
wrote:
The filament is overwhelmingly resitive so the current and votage must
be in phase.
Even if the current is not switched on until it's 90 degrees out of
phase ?
I think you are getting a little confused here. Sketch the voltage
waveform produced by a dimmer. Now sketch the current waveform you
would expect if that voltage waveform went into a purely resistive
load. Hint: I(t) = V(t)/R.


Well even that's not right, dI/dt cannot be infinite and I is at zero
to start with, and also every time the "perfect switch" closes every
half cycle.

Its VERY fast..thats why you have an indiuctor in there to slow it down..


That aside, if you were to carry out a fourier series analysis on the
resultant I waveform you would find it has a strong component at the
50 Hz. fundamental (plus plenty of harmonics) and a constant delay
term related to the phase angle at which the "perfect switch" closes.

It is this phase angle which governs the power transmitted by the
dimmer, and also causes the current to lag the voltage.

I do restrict the discussion to cheap, simple phase angle devices such
as lamp dimmers and drill speed contollers. If there are "true
sinewave" dimmers (as there are "true sinewave" inverters.) I've not
come across any, and probably couldn't afford them if I did.

They exist all right. Fr motor control. Complex PWM plus a smoothing
inductor to iron out the ripple in each phase..

DG

Chris Charles wrote:
That aside, if you were to carry out a fourier series analysis on the
resultant I waveform you would find it has a strong component at the
50 Hz. fundamental (plus plenty of harmonics) and a constant delay
term related to the phase angle at which the "perfect switch" closes.

It is this phase angle which governs the power transmitted by the
dimmer, and also causes the current to lag the voltage.

I do restrict the discussion to cheap, simple phase angle devices such
as lamp dimmers and drill speed contollers. If there are "true
sinewave" dimmers (as there are "true sinewave" inverters.) I've not
come across any, and probably couldn't afford them if I did.

DG

Nonsense


You cannot have a power factor with a resistive load. By definition.


It depends entirely on how you define power factor.

The concept becomes almost meanibngless when applied to non linear
loads, which a dimmer in series with a resistor, is.

And in fact the current slightly leads the voltage in a normal bulb with
no dimmer anyway as it heats and cools through the cycle..

The current and voltage are always in phase.

You should have gone a little beyond O level Physics.

Andy Hall wrote:
On 2008-03-10 10:36:48 +0000, The Real Doctor
said:

On 6 Mar, 13:45, David Hansen wrote:

Nice try

Yes! We used to count David's "Nice try" postings over on uk.railway,
but gave up when he got to several hundred, Does he do "Excellent.
Personal abuse." here as well?

Bless him.

Ian

Oh, he's not a trainspotter as well, is he?


"If all the world ran on bicycles, real ale, beards, anoraks and
electric trains; we could all sing folk songs and be a proper communitee'

In article ,
The Real Doctor writes:

The electricity companies get very cross if one strays too far from a
power factor of 1.0 on a domestic meter. My recollection is that they
used to allow up to 0.9 lagging and no lead at all, but I am open to
correction.

Domestic meters neither know, nor care what the power factor is.
They measure the true power (or more correctly, the energy used,
the integral of power).

--
Andrew Gabriel
[email address is not usable -- followup in the newsgroup]

On 11 Mar, 00:00, (Andrew Gabriel) wrote:
In article ,
The Real Doctor writes:



The electricity companies get very cross if one strays too far from a
power factor of 1.0 on a domestic meter. My recollection is that they
used to allow up to 0.9 lagging and no lead at all, but I am open to
correction.

Domestic meters neither know, nor care what the power factor is.
They measure the true power (or more correctly, the energy used,
the integral of power).

Not true. Unless you think that attaching a suitable reactive load in
order to run the meter backwards means that your house is suddenly
producing energy ...

But quite apart from that, it's not just a question of power. The
infrastructure for supplying reactive amps costs just as much as the
one for resistive amps, and if you hand them back unused no power
meter will see 'em. Hence commercial users with large reactive amp
requirements negotiate individual deals.


Ian

On Mon, 10 Mar 2008 22:09:03 +0000, Derek Geldard
wrote:

On Mon, 10 Mar 2008 20:41:51 +0000, Chris Charles
wrote:

On Mon, 10 Mar 2008 19:48:38 +0000, Derek Geldard

I do restrict the discussion to cheap, simple phase angle devices such
as lamp dimmers and drill speed contollers. If there are "true
sinewave" dimmers (as there are "true sinewave" inverters.) I've not
come across any, and probably couldn't afford them if I did.

DG

Nonsense


You cannot have a power factor with a resistive load. By definition.


Your definition does not encompass active components.

The current and voltage are always in phase.

Clearly, nay obviously, they are not.


Ok, if that is so what is producing the phase shift.

It's implicit in there being a delay between the voltage impressed and
the current drawn.

DG


Nothing can be brushed off as implicit unless definitions exist. What
are your definitions? Why are you saying there is a delay between the
voltage and current?

It is implicit that the current through a resistive load is in phase
with the applied voltage. This is based on the definitions of
resistance and reactance and their voltage/current relationships This
is proved in theory and practice. Nothing can make it otherwise unless
the laws of physics are changed, or new ones defined.

If the load has inductance and/or capacitance present then some phase
shift will occur but it will be negligible if the load is
predominately resistive. This is the case with a filament lamp.

The basic dimmer units are only gates/switches and all that happens is
that they are conducting for part of the cyle. As soon as they conduct
a resistive load draws current, with no delay. Where does phase shift
enter into it?

The effect on the supply is another matter as the current is drawn for
only part of the cycle , which may not go down well with the
suppliers, but it is not phase shift as v and a are still in phase at
your terminals.

On Mon, 10 Mar 2008 23:46:26 +0000, The Natural Philosopher
wrote:

Chris Charles wrote:
That aside, if you were to carry out a fourier series analysis on the
resultant I waveform you would find it has a strong component at the
50 Hz. fundamental (plus plenty of harmonics) and a constant delay
term related to the phase angle at which the "perfect switch" closes.

It is this phase angle which governs the power transmitted by the
dimmer, and also causes the current to lag the voltage.

I do restrict the discussion to cheap, simple phase angle devices such
as lamp dimmers and drill speed contollers. If there are "true
sinewave" dimmers (as there are "true sinewave" inverters.) I've not
come across any, and probably couldn't afford them if I did.

DG

Nonsense


You cannot have a power factor with a resistive load. By definition.


It depends entirely on how you define power factor.

The concept becomes almost meanibngless when applied to non linear
loads, which a dimmer in series with a resistor, is.

And in fact the current slightly leads the voltage in a normal bulb with
no dimmer anyway as it heats and cools through the cycle..

The current and voltage are always in phase.

You should have gone a little beyond O level Physics.

Wit respect - Cobblers!

I'm not defining power factor - the definition already exists.

Varying the resistance doesn't alter the phase - as soon as the
resistance changes the current does so as well.

Phase relationships are the same at whatever level they are studied.
The more complex the circuit the more effects have to be taken into
account but these are interactions of the basic building blocks, not
completely new concepts.

On Tue, 11 Mar 2008 01:27:55 -0700 (PDT), The Real Doctor
wrote:

Not true. Unless you think that attaching a suitable reactive load in
order to run the meter backwards means that your house is suddenly
producing energy ...

I'd think that's pretty unlikely. For one, power meters of the spinning-wheel
type do deal with the common on garden reactive loads very well and accurately
over a long time. And for another, ISTR that power meters can't ever run
backwards, 'cause they have a mechanical widget to prevent just that. A fully
qualified electrician, breaking the seals and making an honest mistake wiring up
a three-phase power meter, which then runs backwards for a few months before
the mistake is corrected is a common story:-)

But quite apart from that, it's not just a question of power. The
infrastructure for supplying reactive amps costs just as much as the
one for resistive amps, and if you hand them back unused no power
meter will see 'em. Hence commercial users with large reactive amp
requirements negotiate individual deals.

Or run their own large banks of capacitors switched to keep the reactive
component within limits...


Thomas Prufer

On Tue, 11 Mar 2008 12:50:42 GMT, wrote:

On 11 Mar,
Chris Charles wrote:

The effect on the supply is another matter as the current is drawn for
only part of the cycle , which may not go down well with the
suppliers, but it is not phase shift as v and a are still in phase at
your terminals.


As you said earlier, RUBBISH!

If the dimmer delays the conduction of the lamp, then the current is delayed.
the voltage is not delayed, so the supply sees an out of phase current
component. thus a less than unity power factor. The load at the output of the
dimmer may be unity power factor, but the dimmer can (and usually does)
affect the power factor seen by the supply.


The voltage to the load is delayed and it is that causes the current
to flow.

In article ,
The Real Doctor writes:
On 11 Mar, 00:00, (Andrew Gabriel) wrote:
In article ,
The Real Doctor writes:
The electricity companies get very cross if one strays too far from a
power factor of 1.0 on a domestic meter. My recollection is that they
used to allow up to 0.9 lagging and no lead at all, but I am open to
correction.

Domestic meters neither know, nor care what the power factor is.
They measure the true power (or more correctly, the energy used,
the integral of power).

Not true. Unless you think that attaching a suitable reactive load in
order to run the meter backwards means that your house is suddenly
producing energy ...

A purely reactive load won't turn the meter at all as it's not using
any energy. In practice, most reactive loads are lossy, so the meter
will record those losses.

But quite apart from that, it's not just a question of power. The
infrastructure for supplying reactive amps costs just as much as the
one for resistive amps, and if you hand them back unused no power
meter will see 'em. Hence commercial users with large reactive amp
requirements negotiate individual deals.

Indeed, which is why I said _Domestic_ meters neither know, nor care.
That's inherent in the way they work.

--
Andrew Gabriel
[email address is not usable -- followup in the newsgroup]

In article ,
Chris Charles writes:

That aside, if you were to carry out a fourier series analysis on the
resultant I waveform you would find it has a strong component at the
50 Hz. fundamental (plus plenty of harmonics) and a constant delay
term related to the phase angle at which the "perfect switch" closes.

It is this phase angle which governs the power transmitted by the
dimmer, and also causes the current to lag the voltage.

I do restrict the discussion to cheap, simple phase angle devices such
as lamp dimmers and drill speed contollers. If there are "true
sinewave" dimmers (as there are "true sinewave" inverters.) I've not
come across any, and probably couldn't afford them if I did.

DG

Nonsense

You are misunderstanding the load.

You cannot have a power factor with a resistive load. By definition.
The current and voltage are always in phase.

Correct. However, we're talking about a resistance in series with
a dimmer (chopper), and this pair doesn't look remotely like a
resistive load.

--
Andrew Gabriel
[email address is not usable -- followup in the newsgroup]

On 11 Mar, 10:44, Chris Charles wrote:

I'm not defining power factor - the definition already exists.

OK, define it for non-sinusoidal voltage and current then.

Ian

On 11 Mar, 19:26, (Andrew Gabriel) wrote:
In article ,
The Real Doctor writes:

Not true. Unless you think that attaching a suitable reactive load in
order to run the meter backwards means that your house is suddenly
producing energy ...

A purely reactive load won't turn the meter at all as it's not using
any energy. In practice, most reactive loads are lossy, so the meter
will record those losses.

There was at one time - ah, kids these days - a common fiddle on the
spinning wheel meters. If you attached the appropriate reactive load,
you could get 'em to go backwards. The electricity board used to get
awful cross if you did this.

Indeed, which is why I said _Domestic_ meters neither know, nor care.
That's inherent in the way they work.

They don't know about power factor, don't care about power factor
and ... in some cases at least ... don't take account of power
factor.

I spent a couple of years of my life developing power dissipation
measrement techniques for highly reactive loads (AC loss testing of
low temperature superconducting coils) and using off-the-shelf power
meters was /never/ an option. In case you're interested, the answer
was accurate calorimetry, calibrated with a resistive load supplied
with DC.

Ian

The Real Doctor wrote:
On 11 Mar, 10:44, Chris Charles wrote:

I'm not defining power factor - the definition already exists.

OK, define it for non-sinusoidal voltage and current then.

Ian

Whats is the angle between a rubber band wound round a stick, and a blob
of plasticene?


Its the phase angle innit? har har.

What is the color of 2.4Ghz? its the powerfactor innit?

The Real Doctor wrote:
On 11 Mar, 19:26, (Andrew Gabriel) wrote:
In article ,
The Real Doctor writes:

Not true. Unless you think that attaching a suitable reactive load in
order to run the meter backwards means that your house is suddenly
producing energy ...
A purely reactive load won't turn the meter at all as it's not using
any energy. In practice, most reactive loads are lossy, so the meter
will record those losses.

There was at one time - ah, kids these days - a common fiddle on the
spinning wheel meters. If you attached the appropriate reactive load,
you could get 'em to go backwards. The electricity board used to get
awful cross if you did this.

I know you could stop them with pin stuck in a hole drilled through the
perspex, bu me and my engineering chums tried all sorts of tricks with
capacitors, and they wouldnt go backwards.


Indeed, which is why I said _Domestic_ meters neither know, nor care.
That's inherent in the way they work.

They don't know about power factor, don't care about power factor
and ... in some cases at least ... don't take account of power
factor.


Oh dear. They are DESIGNED to take care of power factor.


I spent a couple of years of my life developing power dissipation
measrement techniques for highly reactive loads (AC loss testing of
low temperature superconducting coils) and using off-the-shelf power
meters was /never/ an option. In case you're interested, the answer
was accurate calorimetry, calibrated with a resistive load supplied
with DC.


Oh dear.

Ian

On 11 Mar, 22:17, The Natural Philosopher wrote:
The Real Doctor wrote:

I spent a couple of years of my life developing power dissipation
measrement techniques for highly reactive loads (AC loss testing of
low temperature superconducting coils) and using off-the-shelf power
meters was /never/ an option. In case you're interested, the answer
was accurate calorimetry, calibrated with a resistive load supplied
with DC.

Oh dear.

Yes, it was quite complicated. Measuring gaseous helium production
rates of a few ml / second is non trivial!

Ian

In article ,
The Real Doctor writes:
On 11 Mar, 10:44, Chris Charles wrote:

I'm not defining power factor - the definition already exists.

OK, define it for non-sinusoidal voltage and current then.

It's exactly the same regardless of waveform...

Power factor = power / (V[rms] * I[rms])

--
Andrew Gabriel
[email address is not usable -- followup in the newsgroup]

On 13 Mar, 00:18, (Andrew Gabriel) wrote:
In article ,
The Real Doctor writes:

OK, define it for non-sinusoidal voltage and current then.

It's exactly the same regardless of waveform...

Power factor = power / (V[rms] * I[rms])

Over what time period are you calculating the rms values? I have a
torch here. It's switched on for about 1 minute per week. When it's
on, it draws 0.3A from its 3V supply. Resistive load. Power factor?

Ian

On Wed, 12 Mar 2008 23:56:40 -0700 (PDT), The Real Doctor
wrote:

On 13 Mar, 00:18, (Andrew Gabriel) wrote:
In article ,
The Real Doctor writes:

OK, define it for non-sinusoidal voltage and current then.

It's exactly the same regardless of waveform...

Power factor = power / (V[rms] * I[rms])

Over what time period are you calculating the rms values? I have a
torch here. It's switched on for about 1 minute per week. When it's
on, it draws 0.3A from its 3V supply. Resistive load. Power factor?

Ian

I assume the torch is powered by a battery (=DC)?

On Wed, 12 Mar 2008 23:56:40 -0700 (PDT), The Real Doctor
wrote:

On 13 Mar, 00:18, (Andrew Gabriel) wrote:
In article ,
The Real Doctor writes:

OK, define it for non-sinusoidal voltage and current then.

It's exactly the same regardless of waveform...

Power factor = power / (V[rms] * I[rms])

Over what time period are you calculating the rms values? I have a
torch here. It's switched on for about 1 minute per week. When it's
on, it draws 0.3A from its 3V supply. Resistive load. Power factor?

Ian


Is this a joke? I hope so . Otherwise where on earth does RMS, power
factor and, by implication ,Phase come into your example?

On 13 Mar, 10:15, John Evans wrote:
On Wed, 12 Mar 2008 23:56:40 -0700 (PDT), The Real Doctor



wrote:
On 13 Mar, 00:18, (Andrew Gabriel) wrote:

Power factor = power / (V[rms] * I[rms])

Over what time period are you calculating the rms values? I have a
torch here. It's switched on for about 1 minute per week. When it's
on, it draws 0.3A from its 3V supply. Resistive load. Power factor?

I assume the torch is powered by a battery (=DC)?

Yup.

Ian

On 13 Mar, 11:21, Richard Forbes wrote:
On Wed, 12 Mar 2008 23:56:40 -0700 (PDT), The Real Doctor
wrote:
On 13 Mar, 00:18, (Andrew Gabriel) wrote:

Power factor = power / (V[rms] * I[rms])

Over what time period are you calculating the rms values? I have a
torch here. It's switched on for about 1 minute per week. When it's
on, it draws 0.3A from its 3V supply. Resistive load. Power factor?

Is this a joke? I hope so . Otherwise where on earth does RMS, power
factor and, by implication ,Phase come into your example?

Ah. You see the problem with the definition offered, then?

Ian

The Real Doctor wrote:
On 13 Mar, 00:18, (Andrew Gabriel) wrote:
In article ,
The Real Doctor writes:

OK, define it for non-sinusoidal voltage and current then.
It's exactly the same regardless of waveform...

Power factor = power / (V[rms] * I[rms])

Over what time period are you calculating the rms values? I have a
torch here. It's switched on for about 1 minute per week. When it's
on, it draws 0.3A from its 3V supply. Resistive load. Power factor?

Ian

Now how do you calculate RMS for a totally non sinusoidal waveform?

NOT easy.

On 13 Mar, 11:39, The Natural Philosopher wrote:
The Real Doctor wrote:
On 13 Mar, 00:18, (Andrew Gabriel) wrote:
In article ,
The Real Doctor writes:

OK, define it for non-sinusoidal voltage and current then.
It's exactly the same regardless of waveform...

Power factor = power / (V[rms] * I[rms])

Over what time period are you calculating the rms values? I have a
torch here. It's switched on for about 1 minute per week. When it's
on, it draws 0.3A from its 3V supply. Resistive load. Power factor?

Now how do you calculate RMS for a totally non sinusoidal waveform?

NOT easy.

If it's periodic it's dead easy : sum of the rms values of the fourier
components (easily proved using the orthogonality of sinusoids).

But that's not the issue here.

Ian

On Thu, 13 Mar 2008 04:30:32 -0700 (PDT), The Real Doctor
wrote:

On 13 Mar, 11:21, Richard Forbes wrote:
On Wed, 12 Mar 2008 23:56:40 -0700 (PDT), The Real Doctor
wrote:
On 13 Mar, 00:18, (Andrew Gabriel) wrote:

Power factor = power / (V[rms] * I[rms])

Over what time period are you calculating the rms values? I have a
torch here. It's switched on for about 1 minute per week. When it's
on, it draws 0.3A from its 3V supply. Resistive load. Power factor?

Is this a joke? I hope so . Otherwise where on earth does RMS, power
factor and, by implication ,Phase come into your example?

Ah. You see the problem with the definition offered, then?

Ian


Please explain yourself.

On 13 Mar, 12:31, Richard Forbes wrote:
On Thu, 13 Mar 2008 04:30:32 -0700 (PDT), The Real Doctor



wrote:
On 13 Mar, 11:21, Richard Forbes wrote:
On Wed, 12 Mar 2008 23:56:40 -0700 (PDT), The Real Doctor
wrote:
On 13 Mar, 00:18, (Andrew Gabriel) wrote:

Power factor = power / (V[rms] * I[rms])

Over what time period are you calculating the rms values? I have a
torch here. It's switched on for about 1 minute per week. When it's
on, it draws 0.3A from its 3V supply. Resistive load. Power factor?

Is this a joke? I hope so . Otherwise where on earth does RMS, power
factor and, by implication ,Phase come into your example?

Ah. You see the problem with the definition offered, then?

Please explain yourself.

It all seems very basic stuff to me. What bit(s) don't you understand?

Ian

On Thu, 13 Mar 2008 05:35:55 -0700 (PDT), The Real Doctor
wrote:

On 13 Mar, 12:31, Richard Forbes wrote:
On Thu, 13 Mar 2008 04:30:32 -0700 (PDT), The Real Doctor



wrote:
On 13 Mar, 11:21, Richard Forbes wrote:
On Wed, 12 Mar 2008 23:56:40 -0700 (PDT), The Real Doctor
wrote:
On 13 Mar, 00:18, (Andrew Gabriel) wrote:

Power factor = power / (V[rms] * I[rms])

Over what time period are you calculating the rms values? I have a
torch here. It's switched on for about 1 minute per week. When it's
on, it draws 0.3A from its 3V supply. Resistive load. Power factor?

Is this a joke? I hope so . Otherwise where on earth does RMS, power
factor and, by implication ,Phase come into your example?

Ah. You see the problem with the definition offered, then?

Please explain yourself.

It all seems very basic stuff to me. What bit(s) don't you understand?


Far more than you, by the look of your contributions to this thread.

No one who knew what they were doing would apply your arguments to the
type of circuits as you do. That is unless they are trying to appear
clever.

Ian

On Mar 13, 11:39 am, The Natural Philosopher wrote:
The Real Doctor wrote:
On 13 Mar, 00:18, (Andrew Gabriel) wrote:
In article ,
The Real Doctor writes:

OK, define it for non-sinusoidal voltage and current then.
It's exactly the same regardless of waveform...

Power factor = power / (V[rms] * I[rms])

Over what time period are you calculating the rms values? I have a
torch here. It's switched on for about 1 minute per week. When it's
on, it draws 0.3A from its 3V supply. Resistive load. Power factor?

Ian

Now how do you calculate RMS for a totally non sinusoidal waveform?

NOT easy.
Really? Why not? The RMS value of a signal over a particular period
is just the square ROOT of the MEAN of the SQUARE of the value under
consideration. Or:

RMS(f(t)) = sqrt(
integral( lowBound := t1, highBound := t2, func := f(t)**2 )
/
(t2-t1)
)

(and that would look a whole lot prettier it I wasn't constrained to
ASCII!)

If the signal is periodic (and one is not trying to be deceptive),
then t1 and t2 should be corresponding points in the period.

If the signal is not periodic, then the one is usually interested in
the limit as the interval tends to practically infinity.


In the case of the torch, the RMS voltage, RMS current, and RMS power
over a week will all be practically zero. However the power factor
will be very close to one.