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

On Aug 3, 9:31 am, bz <bz+...@xxxxxxxxxxxxxxxxxxxx> wrote:

<A bit of house cleaning and reordering>

[You multiplied by 0.707 instead of dividing and I forgot to multiply by 2
pi]
Agreed?

Yes all of the above and also some confusion about
peak and peak-to-peak. 1.414 and 0.707 and other twofers. >:-)

http://upload.wikimedia.org/wikipedia/en/thumb/1/17/ACPower03CJC.png/300px-ACPower03CJC.png
Second figure from:
http://en.wikipedia.org/wiki/Power_factor

X_C = 120 ohms
R = 0.010 ohms
rms = 120v
pp =170 v
peak = 85 v (4.5 divisions on wiki scope)
75.5 volts inst at 45 degrees into cycle is 4 divisions

1 amp inst at 45 degree mark is my guess assuming 1/2 amp per
division

But bz comments:
<< it is NOT I (RMS current) and E(RMS voltage) that you must use >>
OK I appended "inst" to show where I tried to read from wiki scope.
Let's see if wiki is showing 500ma/division.
I = E / Z
I = 120 volts / 120 ohms
I = 1 ampere (you said 0.95)
An 85 volt inst peak would result in a 1.4 amp inst peak
90 degrees later That is 3.5 divisions on the wiki scope.
Very close to 500ma per division?
Regarding the "Average power" line at y = 0 and the
2f component. A resistor will get just as hot if you swap
the plus and minus connections to its DC supply so the
display seems correct.

--Integration and absoute value signs--
Bz wrote:
<< 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. ** >>

OK I agree 100 watts peak at 45 degrees on the scope
which is 90 degrees at 120 Hz.

Let's look at the power trace at 135 degress. That is a negative peak
I think this is where I misunderstood how you were integrating.
The curve is symetrical in the second picture because it is pure
reactance.
The curve is offset in the third picture because there is some
resistive load.
It is clear to me now that your sampling technique can derive the
120Hz power trace. What I am a bit unsure of, is whether averaging
that over a cycle removes the apparent power. I'll tinker with some
more reasonable values with phi 45 degrees and see if something
lights up because it is still troubling if we calculate 10 watts or
better for a load that can have a peak power of only 10mW.
It is surely the apparent power causing the error but I'd like
to be able to show it with the sampling technique you
are using.

I am a bit tired for that now.
I'll save what I snipped in case I missed some pertainent issues.
Reading from the same *** music resolves most of them I think

Sue...


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