Re: Re Secondary Resonance LC and Harmonics

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Mike" <mikev@xxxxxxxxxxxxxxxx>

Hi,
         A few comments, As any ham or FM broadcast CE knows, using strip
line, plumbers delight construction in an RF amplifier tank or harmonic
filter, link coupled, straight lines work just fine, but we already know
this.
More interesting is the question Phil asked at the end of this posting,
along with the statement that prompted the question:
" >  > Furthermore classic description also predicts that exactly half of
the
>  > inductance of a straight wire resides within the wire itself.
>
>      I don't understand this statement at all. How do we observe this
> hidden inductance? How does inductance "reside" somewhere? Where does the
> other half of the inductance "reside"?
>
> -Phil LaBudde
If this is to be seen, maybe with neon tubing bent as a coil or as part of a
coil and with glass fittings to which a vacuum pump is attached. If high
frequency is to prefer flowing in one part Vs another part, this would show
as glow. In my work with the glow discharge tube, I can tell where power is
flowing greater and where it is flowing less. I can also reduce the pressure
and make the discharge more constricted, dime sized to the entire width of
the 22.5 inch ID of the tube.
If you do this with a chamber, you can see what is flowing within the glass
tubing and perhaps what is flowing outside of it. Electrodes would be within
the neon sign tubing and if there is a second filed, that would ionize
outside the tubing, possibly making this point or not making it.
Regarding the displacement current, we could show this with dust maybe, much
like one can use iron dust on a paper over a magnet to show the shape of the
flux lines.
Just a thought.
Sometimes when a question comes up like "How do we see this", a good one,
it's almost more interesting to figure how to see it rather than if what we
are looking for is even there.
I know I often refer back to using glowing gas but it's a tool I happen to
be well equipped with to use.
Several other people have access to vacuum equipment on the list, too.
The ionized gas in the neon sign becomes the conductor for this test, the
worst part is getting a glass blower to make the turns and splice the raw
tubing sections together if you want a large coil.
There is no reason that an entire secondary for  TC not be made of evacuated
sign glass; I guess that in itself would be something rather cool to see.
Well insulated turn to turn, too!
So maybe this test can be done.
Let the .EDU guys pay for the glass, it's their statement to defend but I
would welcome the result.
Mike


----- Original Message -----
From: "Tesla list" <tesla@xxxxxxxxxx>
To: <tesla@xxxxxxxxxx>
Sent: Thursday, June 30, 2005 8:12 PM
Subject: Re: Re Secondary Resonance LC and Harmonics


> Original poster: FIFTYGUY@xxxxxxx
>
> In a message dated 6/29/05 5:49:15 PM Eastern Daylight Time,
> tesla@xxxxxxxxxx writes:
>
> We can examine the classic equation for an air cored inductor: L = u
> Nsqrd A / l
> Multiply the numerator and denominator by 4pi  we get:
>
> L = u Nsqrd 4pi sqrd  r sqrd / 4 pi l  =  u (2pi r N)sqrd / 4 pi l
>
> Since 2pi rN equals wire length we can write:
>
> L = u (wire length)sqrd / 4 pi l
>
> Letting the solenoid height ( l ) equal the wire length,  we find
> that the classic equation predicts that the inductance of a straight
> wire is simply:
>
> Lstwire = u wire length / 4 pi
>
>      But using the above "classic equation" also collapses the inductance
> to zero as the radius "r" and the turns "N" go to zero.
>
> -Phil LaBudde
>
>  >True if R and N are both zero then we have no wire and hence no
>  >inductance.
>
>      My intent was to show quite the opposite: If R and N are both zero
> then we have a STRAIGHT wire with minimum inductance. Actually, minimum
> inductance is obtained by doubling the wire back on itself so one half
> cancels the other's inductance.
>
>  > I think what you meant to ask is what happens as R
>  > approaches zero for a constant length of wire.
>
>      Sure...
>
>  >As R approaches zero the solenoid length (l) approaches the wire
>  >length. The substitution we used where (l)= wirelength does not lead
>  >to a contradiction, it should be a legitimate step.
>  >To help wrap your mind around this geometry consider what happens when
>  >you take wire from a fat inductor and wind it around progessivly
>  >skinnier forms. Sure enough as the radius gets smaller the length (l)
>  >aproaches that of the straight wire.
>
>      With you so far.
>
>
>  >Of course the really interesting implication is that at R=0, N is
>  >infinite.
>
>     I disagree. Because we have not changed the thickness of the wire, as
> R=0, at some point we must GIVE UP turns to lay the wire on top of itself.
> At some very small R, the wire must physically "corkscrew" and have spaced
> turns in order to complete the turns. In this way, we have used the same
> wire to form less turns.
>     Which brings me back to my original point: as both R and N go to zero,
> so does the inductance. The only way you get a straight wire (R=0) to have
> any turns ("N") is via eddy currents.
>
>
>  >Jared and Larry
>
>      Also:
>
>  > Furthermore classic description also predicts that exactly half of the
>  > inductance of a straight wire resides within the wire itself.
>
>      I don't understand this statement at all. How do we observe this
> hidden inductance? How does inductance "reside" somewhere? Where does the
> other half of the inductance "reside"?
>
> -Phil LaBudde
>
>
>
>