RE: (Fwd) RE: Longitudinal Waves

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Pete Komen by way of Terry Fritz <twftesla-at-qwest-dot-net>" <pkomen-at-zianet-dot-com>

Hello Dave,

I was describing ringdown of a tuned circuit without added energy.  Energy
could be added using a pulse or by feeding the circuit with a sine wave at
the resonant frequency.  Pulse fed (I am not sure where the pulse would go
for best efficiency or highest voltage or ?), you would see the pulse (if
you watch the voltage on a scope).  Sine wave fed you would see a gain in
amplitude unless losses are larger than the energy added, but a smooth sine
wave (kind of the reverse of a ringdown).

Watching an AC voltage supply on a 'scope won't show any bumps or damping;
just a nice fixed-amplitude, fixed-wavelength sine wave.  That's what you
get from the power company.

I think Terry has some pictures of oscilloscope traces that show voltage of
a TC primary and other stuff.

Regards,

Pete Komen


http://hot-streamer-dot-com/TeslaCoils/MyPapers/MyPapers.htm

-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Friday, February 15, 2002 6:41 PM
To: tesla-at-pupman-dot-com
Subject: RE: (Fwd) RE: Longitudinal Waves

Original poster: "David Thomson by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<dave-at-volantis-dot-org>

Hi Pete,

Your explanation is helpful.  At what point of the cycle is new energy
added?  We know there are losses in the system, so at some point the energy
has to be replaced, correct?

I'm going to jump ahead of your answer, because I can't see how energy would
be efficiently added gradually through the entire cycle, it must be added as
a pulse at a given time.

Most likely, just a guess, this energy will be added shortly after the
magnetic field begins to collapse in either one or both places in the cycle.

If the energy is added in just one point of the cycle, then due to the
gradual decay in the sine wave (resistance) there will be a slightly higher
voltage in the cycle just after the added pulse than just before the added
pulse.

Am I correct?

Whether my assumption is correct or not, that is what I had intended to
convey the first time.

Now when I see a perfect sine wave floating across the screen, and I know
due to the laws of nature that there must be resistance in the circuit, I
should be seeing a slight bump somewhere in the sine wave.  But I don't see
it.  Was it smoothed out by the oscilloscope?

Dave

-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Friday, February 15, 2002 12:15 PM
To: tesla-at-pupman-dot-com
Subject: RE: (Fwd) RE: Longitudinal Waves


Original poster: "Pete Komen by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<pkomen-at-zianet-dot-com>

Dave,

Consider a circuit consisting of a capacitor (F) and an inductor (H)
connected in series with nothing else in the circuit.  Suppose that the cap
is charged to some voltage V (if you want, imagine a switch in the circuit).
The cap has charge Q = F * V and energy J = F * V^2 / 2.  When the switch is
closed the voltage is at a maximum and rapidly falls (based on the resonant
frequency of the circuit) to zero as the current rises to its maximum.  At
this point (1/4 through one cycle), the energy is all stored in the inductor
J = I^2 * H / 2 (I = current)

V = inductance * dI/dt (voltage is inductance in Henries * change in current
With Respect To change in time.  Note that when voltage is zero, the current
is at a maximum but not changing (instantaneous)

At this time the inductor starts to give up energy as voltage builds in the
cap and the current changes more and more rapidly (falling) as the current
flow is opposed by the voltage building in the cap.  Finally, all energy is
transferred to the cap (the voltage is reversed from the beginning) and the
current is zero.  This is the end of the first half cycle.  This continues
until all energy is lost.

Energy is lost in resistance and radiation (is there anything else?).  The
energy lost means a lower voltage in the cap when the voltage is at maximum,
and a lower current when current is at maximum.

If it is damped, the Q of the circuit is low and drop in voltage at each
peak is less. (or there may be no second peak).

As usual, I don't know if this is what you were looking for.

Regards,

Pete Komen

-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Wednesday, February 13, 2002 4:51 PM
To: tesla-at-pupman-dot-com
Subject: RE: (Fwd) RE: Longitudinal Waves

Original poster: "David Thomson by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<dave-at-volantis-dot-org>

Hi Malcolm,

>Having spent half a lifetime repairing oscilloscopes as well as calibrating
them I must strongly disagree with the statements in that paragraph. Bear in
mind that you can invert one channel of most if not all scopes. The fact
that there is no DC shift in a pure sinusoidal waveform when you do that
speaks volumes.

I've been looking for the reference that I derived my information from.  I
cannot find it just yet.

The DC shift has been shown by some experimenters to exist.

The shift was explained just as I am presenting it, that there are two
opposite polarity waves working together.

While we are on the topic, what is the current scientific explanation for
the variation of voltage in a damped sine wave?  What force is believed to
determine when the wave will reverse direction?  I could use a little
education on the current theory if you would indulge me.  I believe this is
useful information for Tesla coils in general.

Dave