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

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]

No... that is not what you propsed above.
phi is imaginary time.

What is imaginary about Phi? Assertion does not make it so. Show me
exactly how an angle is imaginary.

It is not a physical *angle*. It is *represented*
as an angle in a polar plot where the waveform
is periodic. You can't turn some
wire 90 degrees and cause inductors to vanish. :o)

No one ever claimed you could.



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
that you can arrive at the correct result without it.


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 in 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.

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)

[...]
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?


[...]
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. 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.


[despirate thrashing and semantic gyrations snipped]

Can the semantics. :o)

Anti semantic?
Proper use of terms is important. You have been misusing 'imaginary'.



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.



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.


http://upload.wikimedia.org/wikipedia/en/thumb/1/17/ACPower03CJC.png/300px-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 )
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.

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

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.


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)




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.

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


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.


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

Semantics again?

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.

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.

[...]

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?


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.

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


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.

Sue...


--
bz

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