Body
Precisely!
On 11/16/2018 10:24 AM, Mick [email protected] [EVGRAY] wrote:
>
> Norm,
>
> Sounds like a Tesla oscillator powered by DC instead of air!
>
> OK need to read links and think...
>
> On 11/16/2018 5:55 AM, Norman Wootan [email protected] [EVGRAY] wrote:
>>
>> Mick, Ole, Kone etal! I have been thinking a lot about my Locked
>> Rotor experiments and did a lot of reading to come up with a fool
>> proof Lenz free motor / generator combination which is so easy to
>> build and proof test! First of all you should read what we know as
>> facts! See the following articles:
>>
>> https://www.motioncontroltips.com/faq-what-are-dc-shunt-motors-and-where-are-they-used/
>>
>> http://www.machineryspaces.com/direct-current-motors.html
>>
>> https://www.quora.com/What-will-happen-if-the-back-EMF-of-a-DC-motor-vanishes
>>
>> https://www.youtube.com/watch?v=3-FPcKgwSzs&feature=youtu.be
>>
>> https://www.youtube.com/watch?v=x3TFALmHtMw&feature=youtu.be
>>
>> https://www.researchgate.net/post/A_DC_shunt_motor_is_running_at_light_load_what_happens_if_the_field_winding_gets_opened
>>
>> Kone loves to play with the very powerful Neos as found at K&G
>> Magnetics and shown in the video of the Ray's LENZ free generator.
>> What I will propose is very simple application of DC motor principles
>> as read from the cited references. First you make your LENZ free
>> motor which is in reality a LINEAR MOTOR (Solenoid) which will give
>> the necessary stroke length to drive the LINEAR generator (Coil
>> oscillating within the drag free, LENZ free area of the Neo magnets).
>> Stroke of the solenoid is controlled by Hall sensor and IGBT drive
>> switch to give an excursion rate (stroke) at a frequency of 60 HZ. By
>> using a DC pulse to drive the solenoid coil it is tricked into being
>> in LOCKED ROTOR condition, thus no LENZ effect. Has the light come
>> on yet?? We know that Ray's generator puts out a sine wave output
>> (AC) with no LENZ drag. Following the DC motor principles as applied
>> to the solenoid driver, you have a motor that is LENZ free also. The
>> combo should give an Over-unity output. Now the tricky part! We know
>> as Ole will agree in that when you switch the solenoid power to off
>> state as magnetic flux is building toward saturation there will be a
>> huge voltage spike (Neutral Spike) which we have to capture and not
>> send to ground as ordinary practice in solid state switch design
>> (snubber diodes) The capture of the SPIKE energy (HV) which occurs
>> naturally will be the over-unity factor here! Interesting concept!
>> Think about it! Norm Mick! (accumulator HV caps)!
>>
>> On 11/16/2018 6:23 AM, Mick [email protected] [EVGRAY] 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, [email protected] [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], <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/
>>>>>
>>>>> Regards
>>>>> Ole
>>>>
>>>
>