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Overunity Machines Forum



Sharing ideas on how to make a more efficent motor using Flyback (MODERATED)

Started by gotoluc, November 10, 2015, 07:11:57 PM

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0 Members and 9 Guests are viewing this topic.

tinman

Quote from: verpies on December 23, 2015, 07:31:31 AM
Yes AFAIK, except he is using a 555 astable to avoid grounding issues.  I think he is working on triggering from the rotor's angular position sensor for the next version.
Already done but not with the latest version.
You are not.

According to the most recent scopeshot:

A) 17.745mJ  Energy expended by the power supply during one ON-pulse, to bring coil's current from 0 to ~600mA,
B)   6.804mJ  Calculated energy stored in the 37.8mH coil at the end of the ON-pulse,
C) 49.005mJ  Recovered energy in C2 after one ON-pulse.



The energy loss in the resistance could've been easily calculated from the RMS value of the current, but I prefer to have the peak current displayed on the scope and Itsu's scope can display only one statistic per channel ...I think.

I divided C by A in order to get the overall efficiency of 276%.    Do you realize what I would've gotten if I had divided C by B ?  ;)

QuoteThe energy loss to the rotor as well as the energy loss due to hysteresis are unknown.

1 simple test would show if any energy is lost to the rotor. Take measurements with the rotor in play,and then take the same measurements without the rotor in play. I think you will find that there will be no energy loss as far as the rotor go's,as i still believe that the energy gain is to do with the external alternating magnetic field's-as i have show in a couple of my test already--although not as precise as you are doing here.

QuoteDo you realize what I would've gotten if I had divided C by B ?

yes = 720.238%

First-an error of 176% just to come out even,is very large. But an error of 620%+ is even larger.
I simply do not see you making this error Verpies,and i also see no other EE guy has stepped up to the plate to help you out. Now if it were me making these calculation,and providing the same outcome's,then im sure i would get run over by EE guy's trying to find the fault in my calculations. But it seems that when some one of your stature and ability comes up with these numbers,all the other EE guys are silent  ::) If there is an error,then it would be an equipment reading error--not yours Verpies.

Just have Itsu run the test without the rotor,and crunch the same numbers. Then you will start to get some answers.


Brad.


verpies

Quote from: verpies on December 21, 2015, 05:20:18 PM
Quote from: partzman on December 21, 2015, 05:15:00 PM
You can use the vertical cursors to define any measurement window in time that you desire with any of the available measurement resources on your scope. It is really powerful IMO!
Show me the cursors re-positioned in such manner, that they encompass only the positive excursion of the red Math waveform and you will make a believer out of me if the Math Mean becomes all positive.
You have made a believer out of me.

Could you help Gotoluc to get his scope screenshots in normal resolution with the statistics on screen ?

verpies

Quote from: tinman on December 23, 2015, 08:38:02 AM
If there is an error,then it would be an equipment reading error--not yours Verpies.
I think Itsu has measured the C2 peak voltage with 2 scopes and verified C2 capacitance with a DMM.

Maybe the CSR is 0.01Ω and not 0.1Ω as expected and we don't know about it, but the input current waveform was also measured by a Hall probe in addition to the CSR and there was not much difference, so Itsu would have to make some systematic measuring mistake here to err.
...like his x10 switch on the probe making a bad contact, or bad probe attenuation setting in the scope ...or other silly thing like that. 
I think I should ask Itsu to do a sanity check and put that scope probe, that measures the voltage from the CSR, across his power supply (or across an AA battery) and see if he gets the same voltage reading as the power supply's voltage (or AA battery).

Anyway, as soon as Itsu makes his "C2 discharger", the C2 voltage waveform will become rectangular with high duty cycle and a regular averaging voltmeter will be able to measure its peak voltage (minus the duty cycle error), so we will have another instrument confirming or denying these measurements.

I was thinking about putting a light bulb in place of R2 because 49.005mJ being recovered 40 times per second, equals 1.96J per second and that is the same as 1.96W of average power.   So a small 2W incandescent light bulb could be lit. 
Who wants to calculate the ideal voltage rating of such small bulb ?

partzman

Quote from: verpies on December 23, 2015, 08:41:11 AM
Show me the cursors re-positioned in such manner, that they encompass only the positive excursion of the red Math waveform and you will make a believer out of me if the Math Mean becomes all positive.

You have made a believer out of me.

Could you help Gotoluc to get his scope screenshots in normal resolution with the statistics on screen ?

Verpies,

I can try! I haven't been following this thread closely so what type of scope does Gotoluc have?

partzman

verpies