Body
Some additional factors that have to be considered such as resonance
both mechanical and electrical. Mechanical is same as in a pendulum
action where there is a natural frequency to target based on weight of
coil and driving mechanism and gravity. (pendulum freq =60 HZ) The
electrical will be accomplished via inductance of coil with proper
capacitive element to achieve the Q factor at operational resonance.
On 11/16/2018 7:55 AM, Norman Wootan 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
>>>
>>
>>