Re: Androcles' logic (was Bilge's logic)

From: Androcles (androc1es_at_nospamblueyonder.co.uk)
Date: 07/20/04 

Date: Tue, 20 Jul 2004 16:16:28 GMT

"Paul B. Andersen" <paul.b.andersen@hia.no> wrote in message
news:cdj34q$1q5$1@dolly.uninett.no...
|
| "Androcles" <androc1es@nospamblueyonder.co.uk> skrev i melding
news:ZoZKc.3729$i13.35416320@news-text.cableinet.net...
| >
| > "Paul B. Andersen" <paul.b.andersen@hia.no> wrote in message
| > news:cdg9qc$3dj$1@dolly.uninett.no...
| > |
| > | "Androcles" <androc1es@nospamblueyonder.co.uk> skrev i melding
| > news:wC8Jc.2866$3Y4.34058262@news-text.cableinet.net...
| > | > This was an accurate observation:
| > | > "They never do, John. Only Paul Andersen is so stupid as to
| > | > try to bluff his way all the way through. I let him run for
| > | > a while, then close the trap. He never sees it coming.
| > |
| > | A - rather arbitrary - example of Androcles closing the trap.
| > |
| > | Paul B. Andersen bluffing:
| > | | But this isn't "in practice", it is a theoretical
| > | | problem specified by YOU.
| > | | Androcles wrote:
| > | | | A------Cap---C----coil----B
| > | | | A voltage V.sin(wt) is applied across A and B.
| > | | | The cap and coil are chosen such that the circuit
| > | | | resonates at w. Assume no resistance in the circuit.
| > | | | What is the voltage across A-C?
| > | |
| > | | YOU, Androcles, specifically said that the circuit
| > | | should be considered to be lossless.
| > | | "Assume no resistance in the circuit".
| > | | You can roll and fart all you want, but the impedance
| > | | of the circuit specified by YOU IS zero at the resonance
| > | | frequency.
| >
| > Correct. Impedance of the circuit specified by me IS zero at the
resonance
| > frequency.
| > See if you can work this one out.
| > Two towns are at sea level with a mountain between them.
| > A battery powered car, fully charged, climbs to the top of the mountain
| > discharging its battery as it does so, and then recharges its battery
on
| > the way down the other side.
| > Assuming no loss, how much energy is left in the battery?
| > Note that the height difference between the towns is zero.
| > Now I wonder if, with that analogy, you can figure out what the charge
on
| > the capacitor might mean in a lossless series LC circuit?
| > Note that I have not specified the height of the mountain or the charge
on
| > the battery, but the journey is lossless.
|
| You cannot change the fact that you are wrong by inventing
| non analogies.

I take it you are saying others that gave the correct response are wrong
too.
If you refuse my assistance then dwell in your own ignorance.

|
| > | |
| > | | You made a giant blunder when you specified
| > | | that a voltage different from zero should be applied
| > | | across an impedance which according to your own
| > | | specification is zero.
| > |
| > | But Androcles called my bluff and closed the trap:
| > | | Not at all. When wt = 0, pi, 2pi... then V.sin(wt) = 0,
| > | | no matter what the value of V. You made a giant blunder, not I.
| > |
| > | And I never saw it coming!
| >
| > I wasn't setting a trap that time, actually. We were debating the phase
| > relationship between B and E, which you claimed was 0 and and I claimed
was
| > 90 degrees.
|
| We weren't "debating" it.
| It's too elementary to "debate".
| Your claim was wrong, and I corrected you.

Assertion carries no weight.

|
| > To illustrate my point, I raised the issue of the lossless
| > series LC circuit, for which you claimed V was zero because the
impedance
| > was zero. That's a little like saying the car can go over the mountain
with
| > an uncharged battery because the height of the towns is zero.
| > John Kennaugh understood it well enough.
| > I still don't think you understand it and are attempting to bluff.
| > Let's ask you outright. Do you still maintain that V.sin(wt) = 0 for all
t?
|
| Yes, of course.

Fine, we are done.
Every engineer that understands an LC series circuit is wong, only Assistant
Professor Andersen is correct.
That's a lot of ego to carry around, nTaul.

|
| > Or to put it another way, is the battery is fully charged at the top of
the
| > mountain?
| > I mean, really.. surely you can see that the current is flowing to drive
the
| > motor as the car climbs the hill, that the battery is flat when it
reaches
| > the top, that the current flows in the opposite direction as the car
| > descends, and the cycle will repeat indefinitely if the system is
lossless?
| > Claiming V = 0 is like saying the mountain isn't there.
|
| I have told you what happens in your circuit before.
| Here it is again:
|
| You made a giant blunder when you specified
| that a voltage different from zero should be applied
| across an impedance which according to your own
| specification is zero.

No other engineer sees it that way.
With a capacitor in the circuit I can raise the DC level to anything I like,
which should be obvious... oops, BLATANTLY obvious to even the meanest
intelligence. At that point, current will cease to flow. A circuit with a
voltage applied through wich noy current flows has an impedance of what,
nTaul?
And if I then reverse the applied voltage, what happens?
Yes, you've guess it. The current will flow and then stop again.
So how do you figure I can only apply a zero voltage?
Because the impedance is zero at the resonant frequency?
The coil and capacitor haven't been replaced by a resistor of zero ohms,
nTaul.
They are still there, just like the mountain in my analogy.
But you carry on in your own belief. Obviously you are ineducable.

|
| What IS possible, however, is to insert a current
| through the lossless series resonance circuit.

And what is the value of that current, nTaul?

|
| (Or you could simply charge the capasitor and
| connect A and B)

Well yes, of course. When you put a DC voltage across A and B, the capactor
charges to the same voltage as A.
A---coil----C-----cap-----B
So the voltage across AC is zero and the voltage across CB is V. Thefore the
voltage AB is V also, and not zero.

| So we are back to my variation of the problem,
| which you were unable to solve:

Not at all.
I presented YOU with MY problem, which YOU were unable to solve and others
did.

| A------Cap---C----coil----B
| A current I.sin(wt) is sent through A-B.

Hence there is a voltage across AB.

| The cap and coil are chosen such that the circuit
| resonates at w. Assume no resistance in the circuit.

Sure.

|
| What is the voltage across A-C?

Infinity.cos(wt)
(real circuits always have some resistance, of course, so we never see that,
but it can get pretty high, enough to blow out my capacitor, which it did.)

| Uac(t) = I.sqrt(L/C).sin(t/sqrt(L.C) - pi/2)
| What is the voltage across C-B?

Infinity.cos(wt+ pi)

| Ucb(t) = I.sqrt(L/C).sin(t/sqrt(L.C) + pi/2)
| What is the voltage across A-B?

V.sin(wt)

| Uab(t) = Uac(t)+Ucb(t) = 0
|
| A rather trivial problem, don't you think?
| But you screw it up.

Yes, nTaul. Of course, nTaul. There there, don't be upset.
Of the course it was the big bad electronic engineer's fault, no
need to cry and stamp your feet and hold your breath. Now
run along to your sandbox and play nicely with the other children.

|
| > |
| > | If you wish, I will give more examples of my stupidity
| >
| > Not my wish. You are free to do as YOU wish, of course.
|
| OK. I will.
|
| Here is a collection of Androcles' traps (read blunders)
| in which he has caught my bluffs (read elementary facts)
|
| Paul:
| | The electric and magnetic fields in an EM-wave are
| | in phase with each other.
|
| Androcles claim they are not in phase.
|
| Blunder!

Well now... some EM waves have really low frequencies.
They were quite common in the early days of radio.
If the E-field, V.sin(wt) and the B field, H.sin(wt) are
both zero at t= 0, pi, 2pi..., where did the energy go?
Into compressing the aether, maybe?
Apparently either Andersen doesn't accept conservation of
energy or he believes in aether.
Where does the energy go at t = pi, nTaul?

|
| Paul:
| | To illustrate this point, I will give a simple analogy.
| | Consider a transmission line terminated with its
| | characteristic impedance R.
| | A sinusoidal wave is propagating on the line.
| | I will assume you know that the voltage and current
| | are in phase in such a wave.
|
| Androcles replies:
| | No, I don't know that. If that is what you think, you are nuts.
|
| Blunder!

Assertion doesn't carry any weight, you know.
Just saying it doesn't make it happen.
Electrical Engineers understand terms like "Power Factor"
that go right over the heads of Assistant Teachers.

|
| Paul:
| | The characteristic impedance of a normal transmission
| | line is resistive.
| | That means that the voltage and current
| | of a travelling wave are in phase.
|
| Androcles replies:
| | Sure. Spout all the nonsense you like. Be a clown.
|
| Blunder!
| The electronic engineer Androcles does obviously
| not know what the "characteristic impedance" of
| a transmission line is! :-)

I'm afraid it is the assistant teacher Andersen that doesn't understand
characteristic impedance, or power factor, or the phase relation
between current and voltage.

|
| Paul:
| | Do you know what is meant by "a 50 ohm coax",
| | Androcles? :-)
| | No?
| | We mean that the characteristic impedance of
| | the coaxial cable in question is 50 ohm.
| | Not 50 ohm capacitive impedance.
| | Not 50 ohm inductive impedance.
| | 50 ohm RESISTIVE impedance.
|
| Androcles replied:
| | I wonder how hot it gets when you apply 250 volts at 50 Hz
| | across that 50 ohm resistance?
| | Let's see... thats err... 5 amps? I^2R = 25*50 = 1250 watts?
| | Ought to be enough to melt the insulator, don't you think?
| | ROFLMAO!
|
| Beyond blunder! He has gone crazy! :-)

Try it, then. Plug your coax into a house socket.
I'll stand well back.

|
| I have a lot more in stock, Androlces.
| I will post them if you keep asking for it.
|
| Paul

Please yourself, nTaul. I may not reply, assertions bore me.
I prefer reasoned debate, which you were once capable of,
but of late you've become another Dinky or Bilge.
Androcles

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