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

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

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.
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.


You have said:
"Note, this is a matter of convenience, NOT a
matter of the reactance actually being imaginary!"

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.

So take phi out of the equation and prove your
statement... or withdraw it.

The dollars in your bank account as calculated by your bank is out of
phase with the cheques you have issued and deposits you have made. That
does NOT make those cheques and deposits imaginary. They are quite
real. Even if a cheque gets lost in the mail, it is real.

All currency is imaginary. The case of Chardonay it
buys is 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.


[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.

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.


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.

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

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

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?


several resistances and reactances can be determined easily. This is
done by representing reactance as a complex number. The real part of
the complex number is associated with the resistance while the
imaginary part stands for reactance.
[unquote]

Note, this is a matter of convenience, NOT a matter of the reactance
actually being imaginary!

You have been laboring under a misconception.

You have been laboring under a similar misconception when thinking
about time in Minkowski space, it is a matter of convenience, not a
matter of time being imaginary!

I realize that it is sometimes difficult to give up ones illusions. But
a scientist is always ready to reexamine ones cherished beliefs and
abandon them when they prove to be at odds with reality. Can you let go
of the idea that an angle is imaginary or show me why it must be?

OK ... I have a capacitor in my hand. What angle do I
turn my wrist so it will behave like an inductor. :-)

Transformations between voltage-time and voltage-phase do not involve
physical rotation of objects, as you certainly understand.

Or would you rather answer how many seconds,
voltage will lag current by, if I connect the capacitor to a
60 Hz supply?

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

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.

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.

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




--
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
.

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