Body
Ole,
So if the cap is disconnected at the even multiple harmonic of the
resonant root frequency synchronized at the proper 1/2 or 1/4 cycle it
is no longer cancelled but carries a charge correct?
On 11/16/2018 9:06 AM, [email protected] [EVGRAY] wrote:
>
>
> Hi Mick,
>
> At resonance (in LC-tank) the capacitive impedance (imaginary part)
> cancels the inductive impedance (imaginary part) leaving only the
> resistance of the LC-tank circuit. The reactance of a capacitor is
> negative while the reactance of an inductor is positive. At resonance
> they are equal in magnitude but of opposite sign which cancel them out.
>
> LC-tank: https://wiki2.org/en/LC_circuit
> Reactance: https://wiki2.org/en/Electrical_reactance
>
> Regards
> Ole
>
>
> ---In [email protected], <mkjekyll@...> wrote :
>
> Ole,
>
> I have had caps charge much faster in an LC tank than otherwise, but
> it could also have to do with noise in the core material as I was no
> using an air inductor.
>
>
> On 11/16/2018 1:19 AM, onielsen@... <mailto:onielsen@...> [EVGRAY] wrote:
>
>>
>>
>> Hi Mick,
>>
>> "If a reactive LC tank capacitor is then disconnected from the
>> reactive source then discharged to a resistive load then
>> reconnected energized back to the standing wave modality
>> recharged disconnected discharged etc. If the reactive energy is
>> harvested in this manner are you claiming the same losses will
>> occur as power factor correcting the reactive back into real power?"
>>
>> Yes that's correct.
>>
>> Common power factor correction in AC is for eliminating the
>> reactive power that doesn't do any work but just heats the wires
>> between the power plant and the reactive load. This is expensive
>> for the owners of the power transmission lines and thus should be
>> avoided. First storing the energy in a capacitor (or inductor or
>> battery) and then dumping it to an active load will still make
>> the power go in just one direction even though it happens in
>> pulses. The energy is then moving from the source to the load.
>> Reactive power has the energy moving to and from the load and
>> power source. The power moving between the capacitor and inductor
>> in an LC-tank is pure reactive except for the loss because of the
>> parasitic resistance in the components and wires.
>>
>> Regards
>> Ole
>>
>>
>> ---In [email protected] <mailto:[email protected]>,
>> <mkjekyll@...> <mailto:mkjekyll@...> wrote :
>>
>> Ole,
>>
>> Disregarding the issue at hand with the patent terminology and
>> not concerning ourselves with the patent just a very high Q tank.
>>
>> If a reactive LC tank capacitor is then disconnected from the
>> reactive source then discharged to a resistive load then
>> reconnected energized back to the standing wave modality
>> recharged disconnected discharged etc. If the reactive energy is
>> harvested in this manner are you claiming the same losses will
>> occur as power factor correcting the reactive back into real power?
>>
>>
>>
>> On 11/15/2018 5:36 PM, onielsen@... <mailto:onielsen@...>
>> [EVGRAY] wrote
>>
>>>
>>>
>>> I have only read to the following paragraph cited below. I
>>> wonder if the applicant knows about what he is writing.
>>>
>>> Citing paragraph 106 of the patent application
>>> (https://patentimages.storage.googleapis.com/53/7d/29/f7b4bd0194992d/US20070296373A1.pdf):
>>> "[0106] There is voltage across the inductor connected in
>>> series to the capacitor and there is voltage across the
>>> inductor connected in parallel to the capacitor. Since
>>> Power=voltage×current, a single power input produces two
>>> branches of electromagnetic power output, increasing the
>>> power output. In the alternative, the present disclosure may
>>> have one electromagnetic power output with less energy input."
>>>
>>>
>>>
>>> It isn't as simple as this. If using an oscilloscope for
>>> measuring the current and voltage it will show some phase
>>> displacement between the current and voltage which means
>>> reactive power. Look at the following small videos to see
>>> how one can be misled if not knowing how the phase between
>>> the current and voltage is: Only the the resistive part
>>> consumes (transforms) energy like R in figure 8.
>>>
>>> 0:42
>>> <https://www.youtube.com/watch?v=yJEZMCCBV3U>
>>>
>>>
>>> Overunity Device Cop 2,5 ? explanation part 3.
>>> <https://www.youtube.com/watch?v=yJEZMCCBV3U>
>>>
>>> 9.5K views5 years ago
>>> 1:01
>>> <https://www.youtube.com/watch?v=Jamj3-2w0eE>
>>>
>>>
>>> Overunity Device Cop 2,5 ? explanation part 2.
>>> <https://www.youtube.com/watch?v=Jamj3-2w0eE>
>>>
>>> 10K views5 years ago
>>> 1:00
>>> <https://www.youtube.com/watch?v=P_E0ck3v40Y>
>>> <https://www.youtube.com/watch?v=P_E0ck3v40Y>
>>>
>>>
>>> Overunity Device Cop 2,5 ? part 1. but,
>>> <https://www.youtube.com/watch?v=P_E0ck3v40Y>
>>>
>>> 8K views5 years ago
>>>
>>> The numbers without the phase shift looks like overunity
>>> which isn't the case when taking into account the phase
>>> shift. Only real power is useful in doing work. Reactive
>>> power doesn't do work except for the small active part
>>> heating the the wires and other components because of their
>>> parasitic resistance. This part caused by current doing work
>>> against the resistance is active power and not actually
>>> reactive power. This is why the power distributing companies
>>> may charge costumers for having high reactive power going
>>> through their cables which heats the power lines without
>>> transferring active power to their costumers. It's just pure
>>> loss because the wires have resistance. Resistance converts
>>> electric power into heat.
>>>
>>> Just looked at the figures and found the description to
>>> figure 46:
>>> "[0149]FIG. 46 illustrates an embodiment of the present
>>> disclosure that conserves light bulb energy. In a test
>>> conducted, on Nov. 16, 1998, a light bulb was installed
>>> according to the schematic disclosed in FIG. 46 and is still
>>> running seven years later.
>>>
>>> [0150]Since Apparent Power (VA)=Current (A)×Voltage (V), the
>>> calculations below illustrate energy conservation for a 60 W
>>> light bulb. The voltage applied at the power input is 115 V.
>>> The apparent power for the Power Input Line (Line PIN) is
>>> measured at 126.5 VA. The apparent power measured for the
>>> shed motor fan is 126.5 VA and the apparent power measured
>>> for the light bulb is 65 VA. Therefore, 126 VA is measured
>>> at the input and a total of 191.5 VA is measured at the
>>> output, indicating an approximately 1.5 gain. Calculations
>>> are also shown for the energy savings for a 75W Bulb."
>>>
>>> It is the apparent power that is increased. The apparent
>>> power is the vector sum of the real power and the reactive
>>> power which makes it always greater than or equal to the
>>> real (or active or true) power.
>>>
>>>
>>> Source with description at:
>>> https://www.allaboutcircuits.com/textbook/alternating-current/chpt-11/true-reactive-and-apparent-power/
>>>
>>>
>>> (Message over 64 KB, truncated)
>>
>