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Interesting. Thanks Norm. Reminds me in part, of free wheeling diodes and collapsing field technology.
On Saturday, 17 November 2018, 1:13:08 am AEDT, Norman Wootan [email protected] [EVGRAY] <[email protected]> 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@... [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
Overunity Device Cop 2,5 ? explanation part 3.
9.5K views5 years ago 1:01
Overunity Device Cop 2,5 ? explanation part 2.
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Overunity Device Cop 2,5 ? part 1. but,
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