On 03/18/2010 12:45 AM, Harry Veeder wrote:

> ----- Original Message ----
>> From: Stephen A. Lawrence <[email protected]>
>> To: [email protected]
>> Sent: Wed, March 17, 2010 2:58:11 PM
>> Subject: Re: [Vo]:add on: OU demonstrated ( with no secrets)
> 
>> Even with the battery hooked up, it would be so easy for the guy
>> to measure current drain from the battery, and measure current into
>> the caps, multiply by the input and output voltage values, and
>> compare the power in and power out -- but he didn't do that;
>> obviously he doesn't want to find out what's really going on.
> 
> He does this near the end of test 6:
> 
> http://www.youtube.com/watch?v=UflGpzijWIA
> 
> and asks for an explanation how 3 volts peak to peak or 562mV rms
> can charge the capacitor to 17 volts.

Does what -- makes simultaneous voltage and current measurements and
determines power output of the battery(s), and the signal generator, and
power input to the capacitors, and compares them?  I watched a good bit
of the video, and skipped around looking for that (it's a bit long
IMHO), but I didn't see anything like it.  I saw him measuring voltages
and I heard him refer to the voltage as being "energy" (which it's not)
but I didn't hear anything about amps, watts, or actual power drain.

Did I miss the power measurement?  Did he actually do it somewhere?  I
don't think he did, but if he did, what did he find?  How did input and
output **power** compare?

As to his question -- how could such a low input voltage (which he
called 'energy', as I recall) produce such a high output voltage -- it's
disingenuous.  He's got a transformer there, duh.

And as to the input -- he's using a heavy duty signal generator in this
video, if my eyes do not deceive me.  Turn it around, you'll probably
find the back is covered with heat sinks.  That's a very serious power
supply there, and again, he hasn't measured the *current* it's putting
out, so he has no idea how much power is being injected into his circuit.

Without knowing input power, the rest is just hot air.  As I already
observed, it's just like running a CF experiment, and claiming excess
heat, but never using a calorimeter, and never measuring input
electrical power.




> 
> note: he simplified the input circuitry from test 4.
> 
> 
> Harry
> 
> 
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