I`m not sure that what is happening here isn`t the old familiar problem you
get with trying to work out the power output of a ham radio transmitter
(particularly ones with valve output stages) when one takes VSWR into
consideration.
To specifics.
Ken wrote:
>This is the site
>that Jean experimented with... he obtained 3.8 Watts without a load
>and 3.6 Watts with a load.
To be more precise, his circuit ***consumed*** more power without a load
and ***consumed*** less power with a load.
In other words he had to put less power in with a load connected than he
did without a load connected.
This is pretty much identical to a radio transmitter when you mistune the
output stage. The extra power consumption usually manifests itself as a
sharp rise in temperature of the output device.
Kem wrote:
>Also notice the voltage potential (stress) in the
>output with/without the load. 1260volts versus 114. That should tellya
>something about impedance mismatches
Precisely - this is exactly what happens in an output stage of a radio
transmitter. Years ago I used to blow up a lot of BFY50 transistors by
failing to take reflected energy effects seriously!
:(
Or to put it all another way: without a load connected, just what was
happening to that input energy? It is 99% probable that he would measure a
rise in temperature of the output transistor / transformer.
The fact it consumes less power when a load is connected just means that
the overall `tuning` of the final `load` (consisting of neon tube and
capacitor) is improved - the output circuit is `fighting` less `reflected
energy` (i.e. standing waves) and so a higher proportion of the input
energy (which itself drops to reflect the increased efficiency) is expended
on doing work `outside` the energising circuit itself - ie. in lighting the
tube.
What is still a complete mystery to me is how logically he moves from
"notice this effect" to "it proves you can `trap` electrons and use them to
reduce energy input".
He seems to just assume that the electronic circuit requires a fixed energy
input to operate at all, and that therefore any reduction from this `norm`
represents the fact of energy coming from elsewhere (free electrons).
Or alternatively, operating at 29kHz, how has he shown that any `extra`
energy (if there was any at all) isn`t coming from VLF transmissions in the
area...etc etc.
Isn`t is just a lot more likely that it is just plain more *efficient* with
a load connected so the power input drops.
Or did I miss something really fundamental???
The only way this experiment could represent anything significant for the
free-energy enterprise is to treat the aperatus as a black box with DC
input (the battery) and DC output (i.e. another transformer and
rectifier/smoothing in place of the neon tube). Measuring DC values is far
safer than trying to work out the notional RMS power values of high
frequency AC signals which are subject to standing wave effects.
Then if the output is greater than the input then he is home and dry.
Nick Hall
Manchester, UK
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