Body
Norm,
I am slowly digesting this but I think my concept of locked rotor was/is
in error. Always thought this was when the rotor is locked to the
rotational field from stator with least amount of slip, seems like I am
reading now it is most amount of slip.
Can you provide what locked rotor means to you and any additional
information of what you think may be occurring beyond the everyday
accepted definition?
Thanks,
Mick
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
>>>
>>
>