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'[EE]:Is this true or is it Bull .... !!'
2002\08\07@064318 by Goring, Steve

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face
I have a vested interest in getting lights to be brighter for less -
Unfortunately I
don't have the money to go to HID lights.

So - is this true or is it a crock !!



Basically what Illustro does is this. Using a standard bulb, be it Tungsten,
Halogen or Xenon (but not HID), a 24 volt pulse is fed in rather than a
constant12 volts. By switching from 0 to 24 volts up to 2,500 times a second
you get a light that is 2.5 times brighter. For example a 60-watt standard
Halogen bulb will give out the equivalent light of a 180-watt bulb. And all
this for the same amount or amperage!



Page can be found at http://www.DIFFLOCK.com - New Products.htm

Thanks very much ..

Steve


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2002\08\07@080548 by Roman Black

flavicon
face
Goring, Steve wrote:
>
> I have a vested interest in getting lights to be brighter for less -
> Unfortunately I
> don't have the money to go to HID lights.
>
> So - is this true or is it a crock !!
>
> Basically what Illustro does is this. Using a standard bulb, be it Tungsten,
> Halogen or Xenon (but not HID), a 24 volt pulse is fed in rather than a
> constant12 volts. By switching from 0 to 24 volts up to 2,500 times a second
> you get a light that is 2.5 times brighter. For example a 60-watt standard
> Halogen bulb will give out the equivalent light of a 180-watt bulb. And all
> this for the same amount or amperage!
>
> Page can be found at http://www.DIFFLOCK.com - New Products.htm


The actual link is here:
http://www.difflock.com/buyersguide/newproducts/illustro/index.shtml

Well it's an interesting idea at least! :o)
The light given off from a heated filament is
dependant on how hot it is. Hotter= more light.

So apparently they are claiming that they can get
the filament hotter using less total power??

AFAIK the filament temp will be determined by the
energy going in vs how quickly it can cool.
I believe that the rate it cools will be relatively
fixed because of mechanical properties of the bulb,
so the brightness *should* be determined totally
by the amount of energy fed into the bulb.

So if they have a circuit that makes a standard
bulb brighter they must be feeding in more power.

In their defense they claim the bulbs are "2.5x the
light output, for the same amount of amperage".
Hmm, they also say "pulsed 24v instead of 12v"

Now I *can* believe their claims of double the light
from a 12v bulb, if they are driving it from 24v with
"same amount of amperage"... ;o)
-Roman

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2002\08\07@081628 by Olin Lathrop

face picon face
> I have a vested interest in getting lights to be brighter for less -
> Unfortunately I
> don't have the money to go to HID lights.
>
> So - is this true or is it a crock !!
>
> Basically what Illustro does is this. Using a standard bulb, be it
Tungsten,
> Halogen or Xenon (but not HID), a 24 volt pulse is fed in rather than a
> constant12 volts. By switching from 0 to 24 volts up to 2,500 times a
second
> you get a light that is 2.5 times brighter. For example a 60-watt standard
> Halogen bulb will give out the equivalent light of a 180-watt bulb. And
all
> this for the same amount or amperage!
>
> Page can be found at http://www.DIFFLOCK.com - New Products.htm

Your link just went to some magazine main page, so I didn't see what these
people are claiming.  However, incandescent bulbs are work on the physics of
black body radiation.  The only way to get more light from the same
fillament is to increase its temperature.  Due to material limitations,
normal incandescent bulbs operate at a lower temperature than that which
would be most efficient for producing visible light.  Therefore, for our
purposes, increased temperature means both brighter light and higher
efficiency.  However, it also means reduced fillament life.  After you work
out all the many contributing factors, bulb life is inversely proportional
to something like the 12th (yes, it really is that high) power of the
voltage.

I don't see how anyone can get around this basic physics without modifying
the bulb.  The filament temperature is a function of the power supplied to
it.  You can certainly drive an incandescent bulb with PWM.  At 2.5KHz, the
fillament temperature is not going to significantly change during a cycle,
so it effectively sees the average power delivered to it, and will emit
light based on the resulting temperature.  The usual brightness versus life
tradeoff should still apply.


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2002\08\07@092407 by Jinx

face picon face
> http://www.difflock.com/buyersguide/newproducts/illustro/index.shtml

"a new set of bulbs need to be fitted at the same time as the unit - The
recommendation is simply because older, well-used bulbs sometimes
give up the ghost the first time that 24volts is passed through them. New
bulbs have no problem with that 'first time experience' "

Bit of a giveaway. My guess is that filaments will get to get be "older,
well-used" a lot quicker with Illustro

And I can't say I particularly care for David Lovejoy's writing. Spelling
mistakes, too many exclamation points, silly comments, and claims like

====================================================

Before we go into details lets look at some facts and figures.

Tungsten

The filament glows a yellow/red colour when a current (limited by law) is
passed through. This system is slightly more effective than mounting
candles to the front of the vehicle"

====================================================

Say what ? There seems to be no direct link to Illustro, just an e-mail
form and very few hits with Google. If buyers find that  Illustro shortens
bulb life it would be very hard to prove, and they'll either not use it any
more or accept that there's a price to pay for brightness

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2002\08\07@102241 by Josh Koffman

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face
It seems a lot like overdriving LEDs with high current, but using
incandescents instead. I don't think the filament will have time to cool
down in between pulses though. I have a little slide rule from Osram
that tells me bulb life relative to voltage. I can't seem to find it
right now (of course) but I seem to recall that a shift of 10% of the
rate voltage can effect bulb life by over 50%. Keep in mind these
figures are from my (very dim) memory, and I believe we decided
automotive bulbs are different then mains bulbs.

Josh
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Roman Black wrote:
{Quote hidden}

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2002\08\07@103251 by Sergio Masci

picon face
----- Original Message -----
From: Roman Black <spam_OUTfastvidTakeThisOuTspamEZY.NET.AU>
To: <.....PICLISTKILLspamspam@spam@MITVMA.MIT.EDU>
Sent: Wednesday, August 07, 2002 1:03 PM
Subject: Re: [EE]:Is this true or is it Bull .... !!


> Goring, Steve wrote:
> >
> > I have a vested interest in getting lights to be brighter for less -
> > Unfortunately I
> > don't have the money to go to HID lights.
> >
> > So - is this true or is it a crock !!
> >
> > Basically what Illustro does is this. Using a standard bulb, be it
Tungsten,
> > Halogen or Xenon (but not HID), a 24 volt pulse is fed in rather than a
> > constant12 volts. By switching from 0 to 24 volts up to 2,500 times a
second
> > you get a light that is 2.5 times brighter. For example a 60-watt
standard
> > Halogen bulb will give out the equivalent light of a 180-watt bulb. And
all
{Quote hidden}

I belive this should read "hotter=bluer light"

>
> So apparently they are claiming that they can get
> the filament hotter using less total power??
>
> AFAIK the filament temp will be determined by the
> energy going in vs how quickly it can cool.
> I believe that the rate it cools will be relatively
> fixed because of mechanical properties of the bulb,

The hotter things are the faster they cool

{Quote hidden}

Regards
Sergio

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2002\08\07@103905 by Michael Simpson

picon face
You can flash a normal bulb at very fast intervals to get a brighter light.
However you cant make a constant light this way or it will burn out the
bulb.

{Original Message removed}

2002\08\07@104912 by Jim

flavicon
face
What's the bulb-life like under this scheme?

RF Jim

----- Original Message -----
From: "Goring, Steve" <Steve.GoringspamKILLspamBSKYB.COM>
To: <.....PICLISTKILLspamspam.....MITVMA.MIT.EDU>
Sent: Wednesday, August 07, 2002 5:42 AM
Subject: [EE]:Is this true or is it Bull .... !!


> I have a vested interest in getting lights to be brighter for less -
> Unfortunately I
> don't have the money to go to HID lights.
>
> So - is this true or is it a crock !!
>
>
>
> Basically what Illustro does is this. Using a standard bulb, be it
Tungsten,
> Halogen or Xenon (but not HID), a 24 volt pulse is fed in rather than a
> constant12 volts. By switching from 0 to 24 volts up to 2,500 times a
second
> you get a light that is 2.5 times brighter. For example a 60-watt standard
> Halogen bulb will give out the equivalent light of a 180-watt bulb. And
all
{Quote hidden}

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2002\08\07@124237 by Brendan Moran

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> I have a vested interest in getting lights to be brighter for less
> - Unfortunately I
> don't have the money to go to HID lights.
>
> So - is this true or is it a crock !!
>
> Basically what Illustro does is this. Using a standard bulb, be it
> Tungsten, Halogen or Xenon (but not HID), a 24 volt pulse is fed in
> rather than a constant12 volts. By switching from 0 to 24 volts up
> to 2,500 times a second you get a light that is 2.5 times brighter.
> For example a 60-watt standard Halogen bulb will give out the
> equivalent light of a 180-watt bulb. And all this for the same
> amount or amperage!

Well, I can explain the principle on which it rests, but as to
whether it is true or not, I cannot tell you for sure.  It seems too
obvious to be in violation of thermodynamics, but it may be.

The principle is not a difficult one.  It is simply a situation where
the RMS power exceeds the average power.  In your example, the
average power is:

(24*.5)^2/R

The RMS power for a 50% duty cycle like you propose is

Prms = (RMS voltage)^2/R
RMS voltage = sqrt(V^2*Duty)

So,

V^2*.5/R = 24^2*.5/R

Now, the comparison between average power and RMS power for this case
is then:

(24*.5)^2 : 24^2*.5
.5^2 : .5
.25 : .5
1 : 2

So you double your effective power.  By the same principle, for any
rectangular wave that goes from 0V to some other voltage, the average
power can be compared to the RMS power by a ratio of

duty% : 1
or
1 : 1/(duty%)

Now, with that principle set, it seems to me that it could not work,
because a system could be set up where a resistor generated heat
using extremely high voltage spikes, with extremely low duty cycle,
ex: 20KV, with a 0.1% duty cycle through a 1 ohm resistor, generating

20000^2/1 * .001 = 400kW effective for (20000*.001)^2 = 400W
Even losses due to ineffective thermal conversion units would pale in
comparison to a factor of 1000.

Thus, while this should work fromt he equations, I don't see how it
can, considering that it would violate the second law of
thermodynamics.

Could someone explain why this doesn't work to me?  Because from the
equations, I don't see why it wouldn't.

My best guess is that the power draw on what ever your power source
happens to be is the RMS power.  Thus you don't gain anything.

Basically, to answer your question, if you put any stock in the
second law of thermodynamics, the principle has to be bull, but, darn
it, it *should* work!

Hope that helps a little...

Now I'll be told to re-take freshman physics, I'm sure.

Dazed and a little confused,
- --Brendan

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2002\08\07@125317 by Olin Lathrop

face picon face
> and I believe we decided
> automotive bulbs are different then mains bulbs.

Oh, and in what way?  And who's this supposed "we" anyway.


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2002\08\07@133413 by Olin Lathrop

face picon face
> The principle is not a difficult one.  It is simply a situation where
> the RMS power exceeds the average power.

Oh no, here we go again with voodoo physics and math!

RMS power is not a meaningful concept.  RMS voltage and current do make
sense, but RMS power, while mathematically possible to compute, has
little relevance to real systems.  RMS voltage and current are useful
precisely because they can determine the *average* power into a resistor.

> In your example, the average power is:
>
> (24*.5)^2/R

Unfortunately you didn't define your example.  I assume the lamp is being
driven with a 0V to 24V square wave.  In that case this formula is wrong.
During the 24V phase the power is:

 P = 24V**2 / R

During the 0V phase the power 0.  The average power is therefore:

 p = 1/2 * 24V**2 / R

which is twice the result of your formula.

> The RMS power for a 50% duty cycle like you propose is
>
> Prms = (RMS voltage)^2/R

No, that is the *average* power.

> RMS voltage = sqrt(V^2*Duty)
> So,
>
> V^2*.5/R = 24^2*.5/R

This is the correct expression for the average power.

> Now, the comparison between average power and RMS power for this case
> is then:
>
> (24*.5)^2 : 24^2*.5
> .5^2 : .5
> .25 : .5
> 1 : 2

This is total nonsense.  Your calculation of "RMS power" is actually average
power, and RMS power is a flawed concept in the first place.

> So you double your effective power.

No, you don't.

> By the same principle, for any
> rectangular wave that goes from 0V to some other voltage, the average
> power can be compared to the RMS power by a ratio of
>
> duty% : 1
> or
> 1 : 1/(duty%)

Nonsense perpetuated.  Going deeper into this gibberish is a waste of time.

> ...

> Now I'll be told to re-take freshman physics, I'm sure.

Basic high school algebra should do in this case.

People will probably say I'm being hard on you, but I don't think so.  I
wouldn't have a problem with this sort of misconception if you weren't
spouting it as fact.  If you are going to profess some knowledge as
definative fact, some people may take it as such.  That's why I just felt
obligated to waste 5 minutes of my life debunking this BS.  If it were left
unchallanged, some newbie might legitmately think it correct.  It's no sin
if math and physics aren't your thing, but then you should label any
assertions properly, like "I'm not sure, but maybe it...".


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2002\08\07@145727 by Josh Koffman

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face
I seem to recall a thread awhile back about bulb life, and how
automotive bulbs last longer than mains bulbs. I think that more than
one person mentioned that the filament was thicker in an automotive
bulb. I couldn't remember who, so I said "we" as in the list in general.

Sorry.

Josh
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fools.
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Olin Lathrop wrote:
>
> > and I believe we decided
> > automotive bulbs are different then mains bulbs.
>
> Oh, and in what way?  And who's this supposed "we" anyway.

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2002\08\07@152038 by Herbert Graf

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face
I remember some of that conversation, I think it was concluded that that was
true of 240V bulbs, they did tend to burn out sooner. However, 120V bulbs
last quite a while, as long as automotive bulbs though? Hard to say,
probably not, and it probably depends on what bulb you speak up.
Instrumentation bulbs usually are rated for something like 17V so they last
FOREVER. TTYL


{Quote hidden}

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2002\08\07@153810 by Olin Lathrop

face picon face
> I seem to recall a thread awhile back about bulb life, and how
> automotive bulbs last longer than mains bulbs. I think that more than
> one person mentioned that the filament was thicker in an automotive
> bulb. I couldn't remember who, so I said "we" as in the list in general.

Automotive bulb fillaments are thicker mostly because the voltage is lower.
Therefore, higher current it needed for the same power.  The thicker
fillament allows for a longer bulb life.  This is the same reason that 220V
houshold bulbs don't last as long as 110V bulbs of the same wattage.


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2002\08\07@175006 by Brendan Moran

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For those of you who simply want to see what I've corrected, just
page down until you don't see ">"s anymore, and do see a
"[RECALCULATION BEGINS HERE]".

I must apoligize in advance for the majority of my post.  I never
wanted to see a discussion go like this.


> > The principle is not a difficult one.  It is simply a situation
> > where the RMS power exceeds the average power.
>
> Oh no, here we go again with voodoo physics and math!

I'm sorry, but that was downright insulting.  As near as I can tell,
all of the math involved is sound.  As to the physics, I apparently
made one mistake at the beginning, which was all you needed to say.
The rest was unnecessary, and does not serve any purpose on this list
other than to embarrass me.

> RMS power is not a meaningful concept.  RMS voltage and current do
> make sense, but RMS power, while mathematically possible to
> compute, has little relevance to real systems.  RMS voltage and
> current are useful precisely because they can determine the
> *average* power into a resistor.

RMS power *IS* a meaningful concept.  It is used to determine the
D.C. power *EQUIVALENT* to the changing waveform power (AC+DC)
dissipated in a load. (ex. the power dissipated by a resistor with
170V peak sine wave applied to it is roughly the same as the power
dissipated by a resistor with 120VDC applied to it). That's the
definition they give you in 1st semester electronics.  Before you
start to think that I'm just some first-year student that doesn't
know anything, let me tell you that I'm not, but how long ago I took
my education is unimportant, simply know that I do have an education
in electronics, and do have some idea of what I'm talking about.

> > In your example, the average power is:
> >
> > (24*.5)^2/R
>
> Unfortunately you didn't define your example.

Sorry about that, I left the quote in at the top, which defined the
example I listed here.  I thought that should have been obvious.

{Quote hidden}

As I said, I made one mistake.  That was basing everything on summing
voltage calculations to start with, rather than summing up the power.
That's all you needed to say.  I would have corrected myself after
that.

{Quote hidden}

Actually, if you follow the math based on the initial equations, it
makes perfect sense.  I will recalculate at the bottom of this post
for your enjoyment.

RMS power is *NOT* a flawed concept.  Perhaps you simply don't
understand it? But you do use it in every day life.  Have you ever
measured the power dissipated by a common light bulb?  I mean *all*
of it.  Heat and light.  I *expect* that you'd find it was not just
the average power, as you seem to indicate.  Of course, I can't be
*sure* since I haven't tried that.  I am, however, relying on what
was taught to me.  On that basis, if you insist that this is wrong,
there are mathematicians and physicists that would do a better job of
explaining to you why RMS is not a "flawed concept" than I can.  I
know one physicist that is quite well accomplished, who I can,
perhaps, get to write an explanation if you wish.

{Quote hidden}

As is my answering this post.  But I'm doing it anyways, because you
seem to have seen fit to attack me personally, rather than just
telling me I made one error at the start.

> > ...
>
> > Now I'll be told to re-take freshman physics, I'm sure.
>
> Basic high school algebra should do in this case.

As I mentioned before, my algebra was perfectly sound.  You may
re-read my original post to confirm that.

> People will probably say I'm being hard on you, but I don't think
> so.

Yeah, I can see people saying that.  You weren't being hard on me.
Just excessive.  If you had simply mentioned the single error, you
wouldn't have wasted so much of your time.  I won't mention how much
of my time you've absorbed for my debunking of your hasty post. (I
say hasty because you claim you wrote all this in 5 minutes.  That is
not enough time for you to properly consider the viewpoint that I was
offering)

> I
> wouldn't have a problem with this sort of misconception if you
> weren't spouting it as fact.  If you are going to profess some
> knowledge as definative fact, some people may take it as such.
> That's why I just felt obligated to waste 5 minutes of my life
> debunking this BS.  If it were left unchallanged, some newbie might
> legitmately think it correct.  It's no sin if math and physics
> aren't your thing, but then you should label any assertions
> properly, like "I'm not sure, but maybe it...".

This should not have been said on-list.  It's not appropriate, but as
you have seen fit to do so, I will respond here.  Anything further
should be kept off-list.

You also spout everything as absolute fact.  If you are going to
complain that others do this, don't do it yourself.  I don't care
what percentage of the time you're right.  That's irrelevant to a
proper code of conduct, which you seem to be claiming that I'm not
following.

You never once provided computational proof that I was wrong in
principle.  Only in calculation.  Incidentally, you never mentioned
the correct formula for RMS power.


Now, as to my earlier error.

[RECALCULATION BEGINS HERE]

Now, then, RMS power is defined as:

sqrt(integral(P^2)/T)
where: P is the power waveform, T is the period of said waveform.

for a rectangular wave,(unless I've made a mistake) that translates
to:

sqrt((Phigh^2*Thigh+Plow^2*Tlow)/T)
where: Phigh is the power of the time that the wave spend in the high
state, Thigh is the time spent by the waveform in the high state each
cycle.  Plow and Tlow are self explanatory given Phigh and Thigh.

Now, since Plow is assumed to be 0, that becomes:

sqrt((Phigh^2*Thigh)/T)

Since Thigh/T is the duty cycle, that is replaced by

sqrt(Phigh^2*duty%)

which gives Phigh*sqrt(duty%)

Now, as Olin mentioned, the average power is:

integral(P)/T

Which in the case of a rectangular wave is:

(Phigh*Thigh+Plow*Tlow)/T

This breaks down to:

Phigh*duty% in a similar manner to the RMS power.

So, the difference between the two is:

RMS:Avg
sqrt(duty%):duty%

RMS is always greater than Avg for all duty cycles less than 100%

A graph of sqrt(x)/x reveals what this relationship actually does for
you.

I'm going to do some more research on the principles behind, and
reasons for the use of RMS as a measurement for power.  If I find
something that I consider to be useful to this group, I will post it.
Where it goes from there is up to the list.

- --Brendan

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2002\08\07@182954 by Harold M Hallikainen

picon face
Sounds like a crock to me! However, I recently read a couple places of a
new kind of filament that keeps radiation other than visible light from
escaping, so the lamp efficiency goes from something like 20% to 60%. I'm
waiting for those!

Harold

On Wed, 7 Aug 2002 11:42:49 +0100 "Goring, Steve"
<EraseMESteve.Goringspam_OUTspamTakeThisOuTBSKYB.COM> writes:
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2002\08\07@183218 by Kevinhoward

picon face
No calculations , no technical values!

If the lamp is prewarmed at (say ) 10% of its rated voltage then the stress
on the filament will be reduced when full or higher voltage is applied thus
lengthening the life of the lamp?
But to what extent of the manufacturers rated life?

Technical calculations could be made as to how much the filament "coil" would
expand and contract at various absolute voltages and the type of metal/s used
for the filament.
Maybe after x thousand cycles the filament would fail due to expansion/
contraction fatigue (along with heat fatigue ) at whatever
temperature/waveform and frequency is presented to the filament.

A Squarewave would provide the harshest environment for an incandescent lamp
0 to X volts in a minimum time- Just look at  the filament! snake in a tube!

Regards

Kevin

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2002\08\07@184714 by Olin Lathrop

face picon face
> As near as I can tell,
> all of the math involved is sound.

No it wasn't, as I thought I pointed out.

> RMS power *IS* a meaningful concept.  It is used to determine the
> D.C. power *EQUIVALENT* to the changing waveform power (AC+DC)
> dissipated in a load.

What you are describing is average power.  The RMS concept is used for AC
voltage and current to find the equivalent DC voltage and current, but this
concept does not apply to power.  The result of those RMS voltage and
current, (or DC voltage and current) calculations is *average* power.

> (ex. the power dissipated by a resistor with
> 170V peak sine wave applied to it is roughly the same as the power
> dissipated by a resistor with 120VDC applied to it). That's the
> definition they give you in 1st semester electronics.

Yes.  Note that you are talking about 170 *volts* peak sine having an RMS
value of 120 *volts*, from which you compute the *average* power.

> As I said, I made one mistake.  That was basing everything on summing
> voltage calculations to start with, rather than summing up the power.
>  That's all you needed to say.  I would have corrected myself after
> that.

Well, you are still going on about "RMS power".

> RMS power is *NOT* a flawed concept.  Perhaps you simply don't
> understand it? But you do use it in every day life.  Have you ever
> measured the power dissipated by a common light bulb?  I mean *all*
> of it.  Heat and light.  I *expect* that you'd find it was not just
> the average power, as you seem to indicate.

I certainly expect it is!

> Of course, I can't be
> *sure* since I haven't tried that.  I am, however, relying on what
> was taught to me.  On that basis, if you insist that this is wrong,
> there are mathematicians and physicists that would do a better job of
> explaining to you why RMS is not a "flawed concept" than I can.  I
> know one physicist that is quite well accomplished, who I can,
> perhaps, get to write an explanation if you wish.

OK, let's do an example.  A 0 to 12V square wave is being driven accross a 3
ohm resistor.  There are several ways to compute the average power.  First
we'll do it by finding the RMS voltage, then using P = V**2 / R to find the
power.

 RMS V = sqrt(12**2 / 2) = 8.485V

That means the resistor is dissipating the same power as with 8.485V DC
applied.  The power is:

 average power = v**2 / R = 8.485**2 / 3 = 72 / 3 = 24 Watts

Now let's find the average power by averaging the power from the two halves
of the waveform.  During the 12V phase, the power is:

 P = V**2 / R = 12**2 / 3 = 48 Watts

During the 0V phase, the power is obviously 0.  Since each phase occurs for
50% of the time, the average power is the average of 48W and 0W yielding 24
Watts.  Note that this is the *average* power dissipated by the resistor.

Now let's examine the RMS of the power signal.  The power signal is a square
wave from 0 to 48W.  The RMS of the power signal is:

 RMS of power = sqrt(48**2 / 2) = 33.94 Watts

As I said, it is possible to compute the RMS of the power waveform, but the
resulting value has no real significance.  If you measure the heat coming
from the resistor, you will get 24 Watts.  If you measure the electrical
power delivered to the resistor, you will get 24 Watts.  The 33.94 Watts
figure has no physical relevance to this example.  It is merely the result
of shuffling some numbers around in a meanigless way.


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2002\08\07@184906 by Spehro Pefhany

picon face
At 02:53 PM 8/7/02 -0700, you wrote:

Raise the voltage, and the efficiency goes up, but the life goes down.

Run a 120V bulb from half-wave rectified 240VAC (RMS voltage is
around 170V) and you get a much brighter and more efficient light bulb
that lasts maybe 10 hours rather than 1500 hours. Ten hours is a
reasonable life for a flashlight bulb, but would be unacceptably
low for many other applications.

Obviously, it takes more power to maintain the filament at a higher
temperature (see Stephan-Boltzmann, it's the fourth power of
temperature considering radiation cooling).

Moving the temperature up increases the amount of energy coming
from the filament that falls into the *visible* range of wavelengths
(see Planck's Law, Wein's Law), which is why we can say that the light
is more efficient. If you count the total energy, including IR and
UV, all incandescent lamps are 100% efficient, of course.

Life decreases exponentially with applied RMS voltage. A trade-off.
By jacking the voltage up on a regular lamp you can make it act
like a short-lived photo-flood lamp with a higher color temperature,
higher efficiency and drastically shorter life, just as you can
reduce it and get long life at much poorer efficiency.

Best regards,

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2002\08\07@194517 by Brendan Moran

flavicon
face
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Hash: SHA1

> > As near as I can tell,
> > all of the math involved is sound.
>
> No it wasn't, as I thought I pointed out.

I'm just going to ignore that.  All mathematical principles were
observed.  I didn't divide one side of the equation and multiply the
other, for example.

> > RMS power *IS* a meaningful concept.  It is used to determine the
> > D.C. power *EQUIVALENT* to the changing waveform power (AC+DC)
> > dissipated in a load.
>
> What you are describing is average power.  The RMS concept is used
> for AC voltage and current to find the equivalent DC voltage and
> current, but this concept does not apply to power.  The result of
> those RMS voltage and current, (or DC voltage and current)
> calculations is *average* power.

I'm beginning to get the feeling that we have a semantic seperation
on this one.  I would view average power as the average voltage
squared over the resistance.  I.e.

4V square wave. 2ohm resistor.

(4V *.5)^2/2 = 2W

where I would see RMS power as AC + DC power.

2W(from the above) + 2V(peak)^2/2ohms
2W + 2W = 4W

I suspect you would call this the average power.  You are saying that
that would be average power, I suspect, because in the case of a
square wave, you happen to get lucky and the full voltage power/2
happens to be the RMS power.

As it happens, the formula I gave also works for this value.

> If you measure the heat coming
> from the resistor, you will get 24 Watts.  If you measure the
> electrical power delivered to the resistor, you will get 24 Watts.
> The 33.94 Watts figure has no physical relevance to this example.
> It is merely the result of shuffling some numbers around in a
> meanigless way.

As I mentioned in the first post, I didn't believe that the equations
I came up with could possibly work due to the second law of
thermodynamics.  Your answer to that was

"Nonsense perpetuated.  Going deeper into this gibberish is a waste
of time."

Well, anyways, I think that the primary stumbling block here has been
a semantic difference.  I view average power as just that: the
average, or DC component.  The sum of the powers divided by the
period of the waveform.  That is what average means.  I see RMS power
as the RMS voltage^2 divided by the load.

What I've said about RMS power being greater than average power holds
true as long as you stick to my definitions (which I hold to be the
correct ones) of RMS and average power.

I think that about clears things up.  Oh, and yes, everyone else I've
ever talked to about this has the same definitions of average and RMS
power as I do.  I guess I shouldn't assume that those are universal.

The rest is moot.

- --Brendan

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2002\08\07@210813 by Russell McMahon

face
flavicon
face
> > Oh no, here we go again with voodoo physics and math!
>
> I'm sorry, but that was downright insulting.

Getting upset by Olin's (or anyone else's) rudeness is pointless and
unproductive.
The best approach is to take the generally sound technical advice, discard
any (in Olin's case, usually but not always small, ) voodoo and gibberish,
ignore all downright insults and general insults and proceed with life.
Don't let 5 minutes of Olin's rudeness (or anyone else's) ruin a whole day
of your time. (Not that I am yet fully adept at taking my own advice :-) ).

Let's try a new approach at the same task.

A Tungsten filament bulb operated at a higher voltage will have more current
flow but not proportionately so as Tungsten bulbs have a positive
temperature coefficient and therefore a higher resistance at higher powers.
This effect is very significant - the inrush current of a cold bulb is many
times that of one at operating temperature.

An Aside:    The only other real contender for bulb filaments is Carbon with
a negative tempco - very hard to make them last). Affects during run up to
temperature will complicate this - if the time on is not large relative to
time to a stable temperature the "Physics" will change somewhat. Cooling
time between cycles will also have some effect on peak temperatures reached
if ton is not large compared to thermal time constant.). I would expect that
thermal tc would be very similar up and down but maybe not (as eg gas
transfer effects in bulb may operate differently )(seems unlikely but good
to think of how things may affect results)

Continuing.

Power while operating is V^2/R.
Energy is Power x time.
Energy is what we care about over a whole cycle.

If resistance at 12v is R then say resistance at 24 V = kR

Let dc be fraction of time on.
At 12v dc = 1

At 24v dc <1

The original info said that the current was the same in both cases.
Here we may well descend into arguments about whether this was average
current, RMS current etc.
We can assume that it was NOT peak current.

I'll assume it was AVERAGE current as I suspect this is what they would
measure. As far as your car battery is concerned, average current is what it
cares about (within reason) for energy purposes. 9Deep discussions on
battery charging, cell chemistries,  gas formation and consequent plate
impedances etc left for another lifetime).

Energy 12v in a unit time = V^2/res x dc
E @ 12v = 12^2/R x 1 = 144/R watt-seconds

For the 24v AVERAGE current to be equal to the 12v average current then if k
= 1
Iavg = Ion x dc                                        - 1
Ion = V/res= 24/res = 24 /(kR)               - 2
Iavg = 12/R (as equal to 12v case)           -3

1,2,3 gives  12/R = 24/kR x dc                 - 4
4 rearranged    dc = 12k/24 (as one might expect)

ie the higher the bulb resistance the longer the bulb must be on at 24v to
make the current equal the current at 12v

E @ 24v  =  V^2/res x dc   in unit time
= 24^2/kR x 12k/24
(k cancels, so why did we bother :-) )

= 288/R watt seconds
ie the energy per unit time is double.

This is an entirely unsurprising result as we deemed that both had the same
average current and as P= V x I then doubling the voltage will double the
power for fixed average current.

Will the bulb be brighter? You bet!
Will the bulb be hotter? You bet!
Will the bulb last less long? Safe bet :-(

Why do it this way at all?
Why not just provide a somewhat higher voltage and run the bulb in a linear
100% on mode - hotter, brighter, whiter, more moth-near-the-flame-ier ?
I don't know
Mayhaps some of the mechanical aspects serendipitously happen to make things
better. Maybe the gas cooling (such as it is, in the bulb. (Bulbs are
typically filled with Argon but I have no idea what % of energy leaves
filament by convection).

All this ignores spectral efficiency of the human eye etc which others have
mentioned. Also moving energy out of infra-red into visible. Also one may
wish to suggest it's NOT energy that counts. Or ...


Now I just know someone's going to criticise me and ruin my day :-)



       RM

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2002\08\07@223123 by Dave Tweed

face
flavicon
face
Brendan Moran <@spam@bmoranKILLspamspamMILLENNIUM.CA> wrote:
> I view average power as just that: the average, or DC component. The sum
> of the powers divided by the period of the waveform. That is what average
> means. I see RMS power as the RMS voltage^2 divided by the load.

If you define your terms rigorously, and work out the math correctly,
you'll find that these are always exactly the same thing.

> I think that about clears things up.  Oh, and yes, everyone else I've
> ever talked to about this has the same definitions of average and RMS
> power as I do.  I guess I shouldn't assume that those are universal.

I guess you've been hanging out with a bad crowd... ;-)

-- Dave Tweed

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2002\08\08@012916 by Sergio Masci

picon face
----- Original Message -----
From: Harold M Hallikainen <KILLspamharoldhallikainenKILLspamspamJUNO.COM>
To: <RemoveMEPICLISTTakeThisOuTspamMITVMA.MIT.EDU>
Sent: Wednesday, August 07, 2002 10:53 PM
Subject: Re: [EE]:Is this true or is it Bull .... !!


> Sounds like a crock to me! However, I recently read a couple places of a
> new kind of filament that keeps radiation other than visible light from
> escaping, so the lamp efficiency goes from something like 20% to 60%. I'm
> waiting for those!
>

I thought a normal household light bulb only converted 4% of the power into
visible light. Can anyone confirm or point to a reference for this 20%
figure.

Regards
Sergio

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2002\08\08@042208 by Wouter van Ooijen

picon face
> RMS power *IS* a meaningful concept.

Sorry, it is not. It is RMS *voltage* that is meaningfull, because it
relates to average power. (because instantaneous voltage squared relates
to instantaneous power).

Wouter van Ooijen

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2002\08\08@050454 by Wouter van Ooijen

picon face
> I'm beginning to get the feeling that we have a semantic seperation
> on this one.  I would view average power as the average voltage
> squared over the resistance.  I.e.

You can view it that way and thus introduce a new concept (average
voltage squared), but it will have little use. Note that RMS means
square first, *then* take average, then take root. You might not like it
(average voltage squared is sure simpler!), but RMS was defined the way
it is for a good reason!

Wouter van Ooijen

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2002\08\08@053321 by Roman Black

flavicon
face
Russell McMahon wrote:

> Why do it this way at all?
> Why not just provide a somewhat higher voltage and run the bulb in a linear
> 100% on mode - hotter, brighter, whiter, more moth-near-the-flame-ier ?
> I don't know
> Mayhaps some of the mechanical aspects serendipitously happen to make things
> better. Maybe the gas cooling (such as it is, in the bulb. (Bulbs are
> typically filled with Argon but I have no idea what % of energy leaves
> filament by convection).


Great post Russell! Here's a thought, what COULD
we do with a SMPS light driver for a vehicle?
* soft starting.
* regulate current exact, independant of vehicle
battery voltage
* measure filament impedance, and know actual
filament temperature, and keep filament at correct
exact temperature.
* hotter filament is more efficient, could overdrive
the bulb as you mentioned, but with enough control
so total bulb life is not too compromised, may even
compare with normal bulbs once you factor in soft
starting and good temp regulation.

The market for these things seems to be 4WD owners
who don't want to spend a lot of money on a whole new
lighting system. Maybe this product DOES give
significantly better performing lights with little
cost in bulb life?
-Roman

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2002\08\08@065903 by Russell McMahon

face
flavicon
face
> The market for these things seems to be 4WD owners
> who don't want to spend a lot of money on a whole new
> lighting system. Maybe this product DOES give
> significantly better performing lights with little
> cost in bulb life?

It would be easy to try. The cost sounded extreme for what it did. A 24 volt
supply with PWM - or add Roman's improvements if desired.  Even easier, use
6V bulbs on 12V *IF* you can still get them (probably rare now). Certainly
easy way to do proof of concept.

A 24V battery supply is easily achieved fro test purposes. Even a Mickey
Mouse (maybe that should be "makeshift" I understand MM means "excellent" in
some domains) arrangement with an extra battery and a double pole switch
could be used. When lights are off connect extra battery across main battery
to charge. When lights are on connect extra battery in series with main
battery. As long as lights on energy need is rather less than extra battery
energy capacity it gives a cheap test bed. But a 150w or so 12 to 24 v
inverter is not too difficult. \



       Russell McMahon

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2002\08\08@071133 by Alan B. Pearce

face picon face
>> RMS power is not a meaningful concept.  RMS voltage and current do
>> make sense, but RMS power, while mathematically possible to
>> compute, has little relevance to real systems.  RMS voltage and
>> current are useful precisely because they can determine the
>> *average* power into a resistor.
>
>RMS power *IS* a meaningful concept.  It is used to determine the
>D.C. power *EQUIVALENT* to the changing waveform power (AC+DC)
>dissipated in a load. (ex. the power dissipated by a resistor with

I have to agree with Olin on this one. RMS Power is not a useful
measurement, and as such does not make sense. It is NOT used to determine
the DC power equivalent. That is determined by taking the product of RMS
voltage and RMS current, while also taking into account any phase angle
shift between the two.

The product of RMS voltage and RMS current is NOT RMS Power. The term RMS
power may be bandied around to signify a power value arrived at by this
product, but that is not really the correct term for what is being measured.

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2002\08\08@084450 by Olin Lathrop

face picon face
> I'm beginning to get the feeling that we have a semantic seperation
> on this one.  I would view average power as the average voltage
> squared over the resistance.  I.e.

This statement is vague.  If you mean

 P = Ave(V**2) / R

then you are right.  However, P is definitely not Ave(V)**2 / R, which is
just a meaningless value.

> 4V square wave. 2ohm resistor.

I will assume you meant 0 to 4V square wave.  The DC component does matter.

> (4V *.5)^2/2 = 2W
>
> where I would see RMS power as AC + DC power.
>
> 2W(from the above) + 2V(peak)^2/2ohms
> 2W + 2W = 4W

This is just plain wrong, not some "semantic seperation".

> I suspect you would call this the average power.  You are saying that
> that would be average power, I suspect, because in the case of a
> square wave, you happen to get lucky and the full voltage power/2
> happens to be the RMS power.
>
> As it happens, the formula I gave also works for this value.

Average power is just that, the average of the power over time.

 Pav = Ave(P) = Ave(V**2 / R)

Since R is a constant, it can be factored out of what is averaged:

 Pav = Ave(V**2) / R

Note that the expression Ave(V**2) is the square of the RMS voltage, or
conversely, Vrms = sqrt(Ave(V**2)).

> As I mentioned in the first post, I didn't believe that the equations
> I came up with could possibly work due to the second law of
> thermodynamics.

I missed that part apparently.  But then why are you clinging to these
incorrect equations?  Let's minimize the incorrect math that gets posted so
that people who aren't sure of these concepts don't get confused.

> Well, anyways, I think that the primary stumbling block here has been
> a semantic difference.

No, these are rigidly defined concepts.  There is no room for your own
interpretation.  You either use the terms correctly or you are just plain
wrong.

> I view average power as just that: the
> average, or DC component.  The sum of the powers divided by the
> period of the waveform.  That is what average means.

Replace "sum of the powers" with "integral of the power" and I agree.  I
think that is actually what you meant, but you need to be careful with
mathematical terms.  See above where I showed this in equations.  Note that
working out the equations for what you just correctly described as average
power comes out to what you now call "RMS power":

> I see RMS power
> as the RMS voltage^2 divided by the load.

Again, see above where I derived this from the definition I think you agree
to for average power.

Once again for the benefit of anyone else listening in that may have gotten
confused by this dicsussion:

 *** RMS POWER IS A MEANINGLESS CONCEPT. ***

> I think that about clears things up.  Oh, and yes, everyone else I've
> ever talked to about this has the same definitions of average and RMS
> power as I do.

There are a lot of erroneous concepts that have gotten perpetuated, and this
notion of "RMS power" is one of them.  That is why I stomp on it loud and
hard whenever it rears its ugly head.  Unfortunately "RMS power" seems to
have stuck with audiophiles confusing people even more.  Even some
manufacturer data sheets use the term, although they are invariably talking
about the average power into a speaker or out of an amp.


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2002\08\08@090541 by Spehro Pefhany

picon face
At 08:40 AM 8/8/02 -0400, you wrote:

>Replace "sum of the powers" with "integral of the power" and I agree.

Specifically, the definite integral over an integer multiple of periods.

A notational nit.. *instantaneous* voltages, currents etc. are normally
represented by lower case letters, whereas average, peak or RMS or DC
values are conventionally represented by upper case letters.

And, yes, I agree with Olin that "RMS power" is meaningless, but it is,
unfortunately, a fairly common way to refer to average or true power IME.

Best regards,

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2002\08\08@092446 by Peter L. Peres

picon face
On Wed, 7 Aug 2002, Olin Lathrop wrote:

>> The principle is not a difficult one.  It is simply a situation where
>> the RMS power exceeds the average power.
>
>Oh no, here we go again with voodoo physics and math!
>
>RMS power is not a meaningful concept.  RMS voltage and current do make
>sense, but RMS power, while mathematically possible to compute, has
>little relevance to real systems.  RMS voltage and current are useful
>precisely because they can determine the *average* power into a resistor.

He meant PMPO power ;-)

Peter

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2002\08\08@102139 by Josh Koffman

flavicon
face
Any mention of how they are achieving this? What manufacturer?

Josh
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Harold M Hallikainen wrote:
>
> Sounds like a crock to me! However, I recently read a couple places of a
> new kind of filament that keeps radiation other than visible light from
> escaping, so the lamp efficiency goes from something like 20% to 60%.

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2002\08\08@162049 by Peter L. Peres

picon face
On Thu, 8 Aug 2002, Roman Black wrote:

{Quote hidden}

* Even hotter plasma, as in HID lamp, since you already have to build a
SMPS upconverter. Just add an ignition circuit and you're almost done. A
HID (or even a Na or Hg lamp) burn at ~50V, it's just the ignition that
takes more volts (~300 for Hg and more for Na).

Peter

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2002\08\09@125042 by Harold M Hallikainen

picon face
On Thu, 8 Aug 2002 06:32:14 +0100 Sergio Masci <KILLspamsmplspamBeGonespamNTLWORLD.COM>
writes:
{Quote hidden}

       Looks like you're right! I was using my faulty memory. Here's an article
on the subject (from the San Francisco Chronicle) that says the new
technique increases efficiency from 5% to almost 100%.
sfgate.com/cgi-bin/article.cgi?file=/gate/archive/2002/06/06/nanot
ch.DTL

       And here's an article from Electronic Engineering Times suggesting an
increase from 5% to 60%.
http://www.eetimes.com/story/OEG20020618S0014


Harold


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2002\08\11@191244 by John

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face
Hello PIC.ers,

Alan said:

>> RMS power is not a meaningful concept.  RMS voltage and current do
>> make sense, but RMS power, while mathematically possible to
>> compute, has little relevance to real systems.  RMS voltage and
>> current are useful precisely because they can determine the
>> *average* power into a resistor.
>
>RMS power *IS* a meaningful concept.  It is used to determine the
>D.C. power *EQUIVALENT* to the changing waveform power (AC+DC)
>dissipated in a load. (ex. the power dissipated by a resistor with

I have to agree with Olin on this one. RMS Power is not a useful
measurement, and as such does not make sense. It is NOT used to determine
the DC power equivalent. That is determined by taking the product of RMS
voltage and RMS current, while also taking into account any phase angle
shift between the two.

The product of RMS voltage and RMS current is NOT RMS Power. The term RMS
power may be bandied around to signify a power value arrived at by this
product, but that is not really the correct term for what is being measured.


>>>>>>>>>>>>>
Then Olin said:


> I'm beginning to get the feeling that we have a semantic seperation
> on this one.  I would view average power as the average voltage
> squared over the resistance.  I.e.

This statement is vague.  If you mean

 P = Ave(V**2) / R

then you are right.  However, P is definitely not Ave(V)**2 / R, which is


<etc, etc. snips>

>>>>>>>>>>

then Spehro said:


>Replace "sum of the powers" with "integral of the power" and I agree.

Specifically, the definite integral over an integer multiple of periods.

A notational nit.. *instantaneous* voltages, currents etc. are normally
represented by lower case letters, whereas average, peak or RMS or DC
values are conventionally represented by upper case letters.

And, yes, I agree with Olin that "RMS power" is meaningless, but it is,
unfortunately, a fairly common way to refer to average or true power IME.


>>>>>>>>>>>>
and now John says:

I'll add my SARands 0.02 cents in here.

When I tried to take issue over this with the `powers that be` at my
former employer over 20 years ago I was given the justification that
`RMS Power' was valid for certain applications.

The job was to size motors and plant for *big* mine winding drives.
These things have ratings up around the 4 - 5 MW area, thus very
long thermal time constants. Their electrical (and mechanical) loading
is quite short-cycle by comparison, hoisting rock up from depth in a
cycle that might take 10-15 minutes.

`RMS Power' requirement was estimated at the mine design stage for a
period which approximated to the thermal time constant of the machine.
The mechanical power requirement was predicted & efficiency taken
into account, giving rise to a strip-chart of electrical power vs. time
over a period of hours.
This was RMS'd to arrive at a motor rating figure.

I never bought the argument that it was more useful than an average
power requirement, but got smitten down. I've a residual feeling that it
gave rise to an over-estimation in the sizing of plant & that this
suited the consulting engineers of the day.
More conservative design, less risky etc.

Never saw a winding motor get a darn bit warm at all...


       best regards,   John

e-mail from the desk of John Sanderson, JS Controls.
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Tel/fax:            Johannesburg  893 4154
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products and services.

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