Re: Intensity and what else affects a single ligth beam's temperature?

"Sue..." <suzysewnshow@xxxxxxxxxxxx> wrote in
news:1186078701.621826.282590@xxxxxxxxxxxxxxxxxxxxxxxxxxx:

On Aug 2, 9:46 am, bz <bz+...@xxxxxxxxxxxxxxxxxxxx> wrote:
"Sue..." <suzysewns...@xxxxxxxxxxxx> wrote
innews:1186048569.997989.74820@xxxxxxxxxxxxxxxxxxxxxxxxxxxx:

On Aug 2, 6:00 am, bz <bz+...@xxxxxxxxxxxxxxxxxxxx> wrote:

{snip stuff I don't doubt you understand]

.....

If the load is pure resistance, we measure the current
now and the voltage now to compute the power now.

If the load is reactive we measure the current now
and the voltage *at some-other-time* (imaginary)
to compute the power now.

NO! you still measure or compute the current NOW and the voltage NOW.
That is exactly what I have been telling you.

Then remove phi from the calculation and demonstate

demon state ?

that you can arrive at the correct result without it.

phi corrects the TIME for v in the following equations so that v and i are
at the same instant in time.
The busses run a bit slow on the 'v line' but that does not make the
passengers on those busses imaginary, nor the time imaginary.

i = Ip sin(wt)
v = Vp sin(wt + phi)
[here i is current, Ip is peak current w is omega, the frequency
factor, t is time
v is voltage, Vp is peak voltage, phi is the phase difference
between
i and v, if any]
....
The instantaneous power is then
p = vi = VpIp sin(wt) sin(wt+ phi)

The current NOW (at this instant of time) time the voltage NOW (at this
instant of time), multiplied together give you the power NOW.



You were going show we don't need imaginary time.
But you used imaginary time in the second equation.

Phi is no more imaginary time than omega is. It just represent the
phase difference between the voltage and the current.

Phi is not NOW so it is imaginary.

Phi is used to compute what the voltage NOW is, starting with the peak
voltage.

We don;t need phi or a calculation to to know the voltage
now, we can read it directly from a scope.

That is what I have been trying to tell you. The voltage and current are
REAL and can be measured in real time. Nothing imaginary about them at all.
If I want to compute the integral of the instant power over a full cycle, I
need a formula to do it, or I need to capture the data and numerically
integrate it. Either will work just fine.


v = Vp sin(wt + phi)

v is the voltage NOW, not at some other time.
i is the current NOW, not at some other time.

p = iv gives the power NOW, not at some other time.

If there is a spark gap or zener it will not perform those
calculations. If there is enough potential NOW.
It will fire NOW... and your scope will show it is NOW.
REAL-LY! :o)

Exactly. Real voltage. Real current. Real power. Instant by instant.


[...]
Show me how Phi is imaginary.

<<...the capacitor's reactance is an imaginary
number (26.5258 -90o, or 0 - j26.5258 ),
the combined effect of the two components will be
an opposition to current equal to the complex sum
of the two numbers. The term for this complex
opposition to current is impedance, its symbol is Z,
and it is also expressed in the unit of ohms, just like
resistance and reactance. In the above example,
the total circuit impedance is:...>>
http://www.allaboutcircuits.com/vol_2/chpt_4/3.html

That does NOT prove that Phi is imaginary.

Most mathematicians have thier hands full proving
that something about their scribblings represents
something real. You want proof that is isn't real?

You have been claiming that certain things are unreal, just because complex
numbers have been used in calculating them.
I have been trying to tell you that they are real. The use of complex
numbers to represent things does NOT make the things imaginary.


[...]
Ok, try this one. The bus line that runs by my house runs a bus every
10 minutes.
Sometimes the busses run late.
Sometimes the busses run early.

I make a plot showing the phase shift of the actual arrival times vs
the scheduled arrival times.
I study the plot and find that the busses are 'capacitive' when traffic
is low and passenger load is low.
Busses are 'inductive' when traffic is high and passenger load is high.

The fact that I use phase and complex numbers in my analysis does NOT
make those riding the bus any less real. It does not make their times
of departure or arrival any less real.

That isn't helpful. AC theory was chosen to illustrate
complex numbers for a good reason.

Yes, but the good reason is NOT that the current, voltage or power is
unreal.
The good reason is that it is convenient for adding phasors to represent
them as complex numbers. That is the ONLY 'good reason'.

Simpler problems
that illustrate it well, just don't seem to exist. Removing
the rungs from a ladder does not make it simpler to climb.
Forgive me if I don't care to board the bus but the problem
is all about a capacitor and the power company playing
ping-pong with some energy. The eyes and jays will
fall into place just like they have for over a hundred years
if we stay focused.

And the instantaneous power changes sign. Integrating over a complete cycle
gives the SAME average power as you get with your formula (converting from
RMS to average).

.....
If 1/60 second is 360 degrees how many seconds is
45 degrees?

1 cycle = 0.017 seconds, so 45 degrees is 2.083E-3 seconds

Sample the current and voltage 45 degrees after a zero crossing

Which zero crossing? Voltage or current? Not that it matters for my
answer.

Lets use current because we have a picture
of that.

Nice picture. It shows why the power waveform has a frequency of 2 times
the voltage waveform. Cute.

It ALSO shows the waveform of the power at each instant, just as I have
been trying to explain to you for quite some time now.

Multiply together. Do you get kilowatts?

That depends on the circuit. You DO get the power. The 'peak' of the
instant power, in the problem you gave, is 15.971 W.

or perhaps 16 pp x 0.707 ~= 10 watts rms.

Close enough, IF we could convert from peak instantaneous power to RMS, but
the peak to peak 200** watts of power averages out to [close to] zero
[0.256 W**] rather than 10 watts RMS. Remember, we MUST integrate over a
complete cycle to get the average power. The power at any instant is like
that steam piston driving that flywheel that I mentioned some time back.

http://upload.wikimedia.org/wikipedia/en/thumb/1/17/ACPower03CJC.png/300p
x-ACPower03CJC.png from
http://en.wikipedia.org/wiki/Power_factor

Look at the first power peak just under the intersection
of a blue and red trace. 45 degrees on the x axis.

I believe that is the point we are discussing.

X_C = 120 ohms
V ~= 120 (the peak is 0.707 x 120 volts )

Wrong. PP = 120 (RMS)* sqrt(2) to get peak from RMS. 120 x 1.414 or
divide by 0.707.
p-p is larger then RMS, not smaller!

To go from p-p to RMS, you _multiply_ by 0.707 (or divide by sqrt(2)).

I ~= 1

Does 1 amp look right for the second graduation on the
y axis and for the current sample at 45 degress?

I get more like 100 watts if I multiply I x E at that point
per your proposal to multiply I x E then integrate.

it is NOT I (RMS current) and E(RMS voltage) that you must use, you must
use i and v (or e), the current and voltage at a particular instant.


Well! I got a tenth of kilowatt; that is almost kilowatts.

Look at the picture again. See the note 'average power zero'? That IS the
integral over a complete cycle.
If we pick the voltage and current at 0.1.97055E-3 seconds (an instant of
peak power), the current is 0.95 A, the voltage is 105.225 V at that
instant in time (Peak voltage is 155.56V, but does not occur at that
moment).
That is going to give 100 Watts at that instant in time. **


Whether it is 16 watts or 100 watts, it seems far from
anything we can call real. 16 watts will keep a cup of
tea warm all day and 100 watts will brew a cup in 1/2 hour.

90 degrees later, the power is minus 100 watts. They sum to zero. The only
power dissipated is in the 0.01 ohm resistor.

Sigh... I really was planning to visit the radio supply for
a few 6SN7's for my calculator. Please excuse the
arithmetic 'till while the filiments are warming. :o)

You want me to send you some 807's? I don't use them so I have some spares.
12AX7s are better for calculators. I built some RS flip flops using them,
back in the early 60's, along with a Heathkit color TV and scope.

That's unreal! You KNOW it is milliwatts or microwatts.
When you skewed the sample for the imaginary current
by an imaginary time (phi) , you got the real answer.

I didn't 'skew' anything. I found the voltage and current at each
instant in time and integrated over a complete cycle.

** I did 'skew up' in that I forgot the 2 pi in my omega, but that does not
make anything unreal.

phi gives you the value at another *time*
That is skewing time.

No. Phi makes sure that we are looking at current and voltage at the same
instant in time. We need it because of how we set up the original
definitions, not because anything is imaginary or unreal.

You need to integrate over one cycle to get the average power.

If you got 16 watts or 10 watts I don't think you got it.
I am getting 100 watts if the current at 45 degress is
about an ampere. That is really missing the mark.

100 watts peak power is correct. I made a mistake in mathcad and forgot to
multiply by 2 pi when feeding in the frequency, w(omega).



Instantaneous power has a meaning but it is not what you pay for.

Semantics again?

No. The power company only charges for the power you KEEP. If you take 100
watts on one quarter cycle and give it back on the next quarter cycle, the
net is zero.

If you don't believe me, still, perhaps you will take this quote
from the book cited above:
[quote from page 85, complex impedance]
It is convenient to represent the two elements of reactance, the
magnitude and phase angle, in such a way that the results of
combining

There is no angle. It is a time.

You are arguing with the book's author now. It IS an angle in this
case. Transforms. You do understand that there is a correspondence
between voltage-time and voltage-frequency or voltage-phase, don't you?
You understand that we can freely transform between representation
systems without changing what actually happens in any way, don't you?

You can use time whether it is periodic or not but the
angular measure only applies for a periodic waveform.
The tail is wagging your dog.

No. If I put any wave form through a reactive component, the product of the
instantaneous voltage and current still gives the instantaneous power.

Transforming from one representation system to another has no effect on
what happens in the real world.
The map is not the territory.
If you forget WHY the map was drawn the way it is, you get the mistaken
idea that Greenland is much larger than it really is.
If you forget WHY the map was drawn the way it is, you get the mistaken
idea that the use of imaginary numbers in calculating current through a
reactive component indicates that the current is not real.
If you forget WHY the map was drawn the way it is, you get the mistaken
idea that movement along the t axis in Minkowski space is not real.


Remember... we WANT some proficiency working
in units of time because Minkowski space won't give
us angles for anything but monocrhomatic light and
that with a bit of gymnastics.

YEP, but that doesn't make the time discrepancy for a moving clock[as seen
from another frame] 'unreal'.

[...]

The voltage will lag the current by 90 degrees (assuming no resistance
or other reactance) or 4.167E-3 s

See! That didn't hurt a bit. Your angle was really a
time wasn't it?

It can be looked at that way, but it is just a model for what happens when
I measure the voltage and current at an instant in time.

I'll take the latter problem. By rejecting a formalism
that labels some things as imaginary, you have lost
track of what is real.

That is totally unreal.

The voltmeter I carry around in my pocket (RS 22-802) does NOT show
instant voltage nor current at all. It does display VDC, VAC[RMS], and
ohms.

I have 2 scopes in my ham shack and a couple more in my storage room.
They WILL show the instantaneous voltage and [with a small series
resistance] the instantaneous current. The A to D input of my laptop
(audio in) can be used to sample the voltage and current on a real time
basis and by using LABVIEW 8.2 (a copy of which I have in front of me
at this moment) I can capture those values and perform the math
necessary to display both the instant power and the average power.

I would suggest using a larger resistor if you are actually going
to sample it. Maybe 20 times the value of the "small series
resistance"
you take the current sample across.

Depends on the sensitivity of the instruments available and the noise
level, but I agree it is easier, in the case under study, across a .1 ohm
resistor than a .01 ohm resistor


If you have 5 ohms, 0.1 ohms and 100uf we
can cheat on the maths.

http://www.allaboutcircuits.com/vol_2/chpt_4/3.html


Nothing imaginary takes place in capturing this data. The data is not
imaginary. The calculations, as I have showed, do not require the use
of imaginary numbers at any point.

Just so the current and voltage samples are simultaneous.

right

(that is the technique you claim will measure real power)

I testify, as an expert witness, that it will allow us to measure real
INSTANTANEOUS power. Which must be integrated over a cycle to obtain
average power which is all we have to pay for.


Unless you can explain to me how my scope or A/D card can access
imaginary things, our conversation is now at an end.

Just so the current and voltage samples are simultaneous.

So, we are in agreement at last and 'even on the number of mistakes'
[You multiplied by 0.707 instead of dividing and I forgot to multiply by 2
pi]
Agreed?

--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

bz+nanae@xxxxxxxxxxxxxxxxxxxx


--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

bz+spr@xxxxxxxxxxxxxxxxxxxx remove ch100-5 to avoid spam trap
.

Relevant Pages

  • Re: Intensity and what else affects a single ligth beams temperature?
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    (sci.physics.relativity)
  • Re: Intensity and what else affects a single ligth beams temperature?
    ... exactly how an angle is imaginary. ... now and the voltage now to compute the power now. ... you still measure or compute the current NOW and the voltage NOW. ... Then remove phi from the calculation and demonstate ...
    (sci.physics.relativity)
  • Re: Intensity and what else affects a single ligth beams temperature?
    ... What is imaginary about Phi? ... now and the voltage now to compute the power now. ... you still measure or compute the current NOW and the voltage NOW. ... matter of the reactance actually being imaginary!" ...
    (sci.physics.relativity)
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