1/4 Wave Theories - Trash Them!
From: terryf-at-verinet-dot-com [SMTP:terryf-at-verinet-dot-com]
Sent: Monday, June 01, 1998 9:08 PM
To: Tesla List
Subject: Re: 1/4 Wave Theories - Trash Them!
I won't attempt to explain 1/4 wave antenna currents here since
there is no way to provide good graphics and such. Any antenna theory book
should provide the details of current and voltage phases on a 1/4 wave
antenna and explain them much better than I. Since this was directed to me
I will add some comments however.
At 01:24 PM 6/1/98 -0500, you wrote:
>From: Barton B. Anderson [SMTP:mopar-at-uswest-dot-net]
>Sent: Saturday, May 30, 1998 2:30 PM
>To: Tesla List
>Subject: Re: 1/4 Wave Theories - Trash Them!
>Terry and All,
>I'm not sold. I know I may get blasted on this, and I hope so if it will
The reason we are all here is test our ideas against the knowledge of
others. Hopefully, the right ones will pop to the surface and the wrong
ones will......... :-)
>But my biggest problem is the definition of the measurements of current
>at the bottom and top of the secondary. In my understanding, in any series
>LC circuit, the phase angle is always 0 degrees regardless of the length of the
>winding, be it a 1/4 wave, 1/2 wave, or everything in between of the *inducing*
>circuit. This is due to the *net* reactance of L and C canceling, leaving
>(effective resistance), and current across a resistive circuit has *no* phase
>variation. It seems from your measurements that you have proven this. But
how does this
>disprove a 1/4 wave effect?
Normal 1/4 wave theory is predicted by the time it takes for light (or
electricity) to travel the length of an antenna. By adding inductance and
capacitance, that length can be artificially shortened. It was this effect
that the 1/4 wavelength theory was based on regarding Tesla coils. You
statements are basically correct for normal inductors and capacitors.
>Maybe, the definition of a 1/4 wave effect is what is mis-represented. All
a 1/4 wave
>effect says to me is that an ac signal has a wavelength (true?)
Yes. However, there are standing wave effects that are also in question too.
>and the ac signal is
>generated from the primary tank circuit of L and C ringing following the
>at the sparkgap (true?).
Basically, although the ringing is as the circuit is completed and stored
energy is allowed to circulate between the L and C.
> ok, every ac signal has high's and low's. The first peak is
>max voltage (true?)
The first "peak" would be a current maximum in the LC circuit after the gap
>and this max voltage will produce max current across R (true?).
1/4Fo latter, the current will drop to zero and the energy in the circuit
will be stored as a maximum voltage peak. R has only minor importance
(assuming it is small). The reactances govern the voltage and current levels.
>the secondary coil has a wavelength (true?)
It has an Fo. Wavelength would be a very qualified value here. Lots of
details to consider....
>and if that wavelength is equal to 1/2 the
>wavelength of the inducing tank, then max voltage should be expected in the
>the coil (true?).
In other words, if the primary were tuned to Fo and the secondary was tuned
to 2Fo. Would you see a voltage peak on the center of the secondary. I
would say no. The secondary would not be tuned to 2Fo and the incoming
energy would be rejected by the mis-tuned LC circuit of the secondary.
There would be some transformer action but it would not be optimal by any means.
>Then also a 1/4 wavelength applied secondary should have an expected
>max voltage at one end (true?).
If the base is grounded, then any voltage maximums that may occur should be
at the top of the coil.
>Then if max voltage is at one end, there *must be* a
>smaller voltage at the other end (true?)
If the base is grounded it would be zero.
>If we multiply the current x voltage at the
>*smaller* voltage end of the secondary, we end up with a x number of watts.
Power(rms) = Vmax * Imax * COS(phase angle) * 1/SQRT(2) Remember that even
though the current and voltage may be high, if they are 90 degrees out of
phase the real power is zero.
>know, power will not increase (for the same time frame) in any system.
If you mean energy, Yes! That is very important to understand!
Instantaneous power can increase dramatically, as in the case of an arc.
>Then if we take
>x number of watts and divide it by the voltage at the *high* voltage end of the
>secondary, we have less current (true?).
The phase angle factor makes this untrue. In a pure LC circuit there is no
>With that said, comparing *phase* of current at the base and top of a coil
with a net
>reactance of 0, will *not* produce a phase variation.
The reactance is not zero. It is 2 * pi * F0 * L or 1 / (2 * pi * C)
The reactance controls the currents and voltages in the system given the
initial stored energy.
>It should however, indicate a
>resonant rise of voltage and current.
"Resonate rise" assumes that there is more energy being pumped into the
system that allows the energy level of the system to be stored and to
increase over time. We only have the initial energy of the system to use.
The maximum voltage will never rise above the initial system voltage (in an
LC circuit). Q only has meaning regarding the system frequency rejection
and such. This is a very important and far reaching conclusion. If we
don't have resonant rise in our coils, the importance of Q changes drastically.
>When I look at the waveforms you captured, I see
>a smaller current at the top than at the base of the coil. Exactly as it
should be. Now
>if you measure voltages instead of currents (if it were possible), you
would see the
>base at a smaller potential than the top (true?).
The base voltage is zero given some small impedances. It is grounded.
Direct field strength (B and E fields) may soon be available.
>Compare a 1/2 wave secondary current measurement to a 1/4 wave secondary.
>circuits tuned to resonance, the net reactance should be 0 producing max
>limited only to Re and the Q of the coil.
If true, It would imply that I could charge the top capacitance to a very
high voltage. The conservation of energy in our system limits the top
voltage to VOmax = VImax * SQRT(Csec/Cpri). Q and Re only have minor
>However, I'll bet you will see a decrease in
>current (top to base comparison) on the 1/4 wave secondary and no current
>the 1/2 wave secondary (again, top to bottom comparison). Move the top
probe to measure
>middle of the coil, and then you will see the current decrease on the 1/2
I am working on equipment that can do this type of "area" probing. Not
available yet. Previous comments apply...........
>I realize that c-top changes the distributed capacity of the secondary, and
>most coils these days run with a large c-top, the 1/4 wave design *may not* be
>advantageous for maximum voltage, but I just don't see how these waveforms
>me that the 1/4 wave theory's are trash? In other words, why are you
looking for a
>phase variation? Did you expect to see one?
Large top terminals may have a big advantage (I will discuss this in the
near future). I did expect to see almost a 90 degree phase shift in the
current in this relatively low loss system that should have produced very
profound standing waves.
>BTW Terry, I enjoy your page, postings, and work with TC's *very much!!!* This
>particular posting I am having trouble understanding how, "what you are
>anything to do with what you are saying".
>Any and All comments *very* welcome,
I hope I haven't just made this all the more confusing. Sometimes
just answering these posts point by point is not helpful. If anyone else
has a better way please feel free. I am not the greatest explainer of
things at times :-)
Thanks for you comments. There are ideas running all over both on
and off this list. Ideas are coming and going very quickly. I hate to
comment an too much until some of the issues are settled down to a
manageable level. Very quickly we should come to a better level of
understanding on all this. There are a number of new and important
principles that seem to be emerging out of all this (like the importance of Q).
This issue is not over, it is just beginning......