Partnered Output Coils - Free Energy

TinselKoala · February 08, 2015, 04:28:48 AM

Same as above but with maximum drive from the 555 oscillator. The HV signal is clipped at the top so the "measurement" in the window is invalid. The neons stay on longer though due to the "double hump" . Raw input current from inline ammeter is 1.08 amps.

TinselKoala · February 08, 2015, 04:58:27 AM

Here's a question for the experts.

My Link DSO will not do trace multiplication so I can't generate an instantaneous power curve. However, under the Measurements, I can select "Area" which will give me the area under a waveform in Volt-nanoseconds.

So... if I ask it to give me the area under the Vout and Vdrop waveforms, then multiply those two values, is this equivalent to multiplying the waveforms and then taking the area (integrating) of that power curve, with a resulting answer in Joules? I don't think the units work out exactly, since the result would seem to be in Watt-seconds² instead of Watt-seconds (Joules)... but anyhow, what do you think? MarkE, Vortex1, poynt99, anyone else who's had calculus?

MarkE · February 08, 2015, 05:00:18 AM

Quote from: TinselKoala on February 08, 2015, 03:51:55 AM
OK, how about this. Let's redefine "output" to be the HV output referenced to "ground" aka the PSU 0 volt rail. Leave the load resistor in circuit because otherwise the L3 coil is just hanging there. Put 10 neons in series with a 4.7 ohm precision non-inductive resistor at HV Out, measure Vdrop across this resistor, and take Vout from TP A and Iout as Vdrop/4.7 . Drive with 555 oscillator with sufficient drive to light neons brilliantly.

And here's what I get. I had to overlap the current (Vdrop) trace on the Vout trace because there just isn't room. The maximum v/div for the Link at 100x probe is only 200 v/div, whereas I can use 500 v/div with the analog scopes.
Input power is 6.7 V and 650 mA (raw from inline meter).

From a noise standpoint I would put the CVR in the low side of the circuit. The power through the cement resistor is already known to be low based on the very slight heating. You could always run a temperature rise experiment to find the equivalent DC power if you can even register it. CE is reporting low mW values.

MarkE · February 08, 2015, 05:01:40 AM

Quote from: EMJunkie on February 08, 2015, 03:49:19 AM
TK, I must say, I did not expect you to through down a list of demands like you have!

I am a bit surprised.

I provided all here and at OUR The Stan Meyer VIC Circuit. Has it not been an exciting experiment?

Chris

What a bummer, replicators actually asking you to be specific about what it is that you supposedly have that they should replicate. Carnival barkers everywhere feel your pain.

TinselKoala · February 08, 2015, 05:04:03 AM

Quote from: MarkE on February 08, 2015, 05:00:18 AM
From a noise standpoint I would put the CVR in the low side of the circuit. The power through the cement resistor is already known to be low based on the very slight heating. You could always run a temperature rise experiment to find the equivalent DC power if you can even register it. CE is reporting low mW values.

Yes, sorry I wasn't clear. The 4R7 is at the bottom of the neon stack, between the lowest neon and the 0 volt (ground) rail.