OU/COP>1 switched cap PS cct like Tesla's 'charge siphoning'

[nul-points]

hi HD

you're right - there does seem to be a slight communication problem!   


in my statement which you quote, there's no mention of an 'extra' amount of joules required to do the work

> Quote from nul-points
"so are we now saying that a physical process which previously required us to do work to overcome the increasing internal polarisation producing the capacitor field has suddenly changed to NOT needing to overcome the increasing internal polarisation, just because an inductor has been added in the external current path?"


at that point i am merely discussing what difference would adding an *inductor* make to the regular charging work function

references to excess energy relate to the results i'm measuring at the resistive load compared to the energy supplied from the input source


> common sense will tell you that the 50% loss is attributed to the non-ideal current source comprised of the voltage source plus charging resistor. This is the culprit and what the textbooks base the "simple charge equation" on. The 50% energy is lost as heat in the resistor, because in this case all of the current must flow through the resistor and it is a lossy device by nature.

i think we're both saying the same thing here


> In your latest test you might want to examine how you arrive at your power dissipated in the resistor and inductor DCR.  The duty cycle is 0.12mS on time, 0.28mS off time and 0.4mS cycle time. It appears you are taking the entire time (9.16 mS) to arrive at your power dissipated in the resistor and inductor DCR. It should be closer to 2.64mS.

i think you're assuming that my calculations don't account for the pulse and cyclic nature of the measured waveform

the voltage waveform across the load resistor is recorded digitally and input to a spreadsheet; the instantaneous power for each reading is calculated; the instantaneous powers are summed for the duration of the relevant section of the waveform; the Sum is divided by the number of cycles to give a value for the Average Power; the energy value is obtained by multiplying the Average Power by the period of the relevant section


> Also it is difficult to compute the actual RMS current per cycle due to the irregular waveform. You might also want to examine the actual power per cycle based on a changing amplitude over 22 cycles.

i think we covered this in the previous answer - the same method is used for obtaining values for all the charging and discharging energy converted by the 10ohm load


there is a way of checking which of us is correct in our assumptions relating to these result calculations:

the final energy discharged from C2 can be found by using the method i just described above - the same energy value stored on C2, prior to discharge, can also be found by noting its terminal voltage and using the regular stored-energy equation

A) Spreadsheet method:
Average Power discharging C2 through 10ohm load: 0.093W
discharge period: 10.44ms
measured Discharge Energy on 10ohm load:  0.093 * 10.44 = 0.97mJoules

B) Stored Terminal Voltage method:
final stored voltage on C2(196uF): 3.12V
stored energy in C2: 0.95mJoules


the two values are in close agreement


the results i posted on my website show both these values

all the best
sandy
________________________________________________________________________________
Doc Ringwood's Free Energy site    http://ringcomps.co.uk/doc    ...bringing you measured Overunity results since May '08

[HEYDUDE]

If what you are saying is true, and you have not made any measurement errors, you should scale this up to the kW range and apply for the OU prize. Should not be hard to do. All energy gains will show up as extra heat and can be easily measured, as the excess energy you claim has nowhere to hide.

Once scaled up  to the kW range you will certainly have the few extra watts to win the prize.

Just design a suitable flyback converter to keep C1 topped up.

Would make a great house heater once scaled up. And if the COP's claimed are true, will beat out pure resistive heaters.

What is keeping you from doing this?

Best of luck in your work.....HD

[Koen1]

@Grumpy: I'm sure you didn't really mean your question like this,
but on the subject of polarising dielectrics, there's a few ways.

A dielectric can obviously be polarised by electrostatic means,
but there's still differences in the type and material phase of
the dielectric. Liquid dielectrics don't polarise in exactly the
same way as solid dielectrics do.
Assuming you're talking about solid dielectric materials, there's
still different ways to polarise those, although it depends on the
material. Some dielectric materials can be melted and polarising
potentials can be applied during the cooling and solidification phase,
which causes the solid dielectric material end product to have a
permanent polarisation. These are generally referred to as "electrets".
Basically these are "self-charging" or "eternally charged" capacitors
(well not exactly but it helps to get the idea across ).

Then, there's also "ferro-electric" dielectric materials. The "ferro-electric"
should be understood as analogous to "ferro-magnetic", in that the
solid and stable material itself gains a specific polarisation when such
a specific polarising charge is applied. Depending on the exact type
of material used, this "ferro-electric" polarisation can be remanent
just like some ferromagnetic metals remain very slightly magnetised
after having been exposed to a magnetic field. However, much like
these magnetic materials, heating or shocking or exposure to
opposing fields will destroy the remanent polarisation and make the
material behave much more like a true dielectric.

And then, there's even "magnetic" dielectrics, which can really be magnetically
"polarised" and the magnetic polarisation and its effects on the structure of the
material have direct influence on the dielectric constant of the material.
These are not generally used for dielectric purposes unless interaction with
magnetic fields is desired, andusually it is not. This is also why not very much
is known about them (outside of very specific research).

So there's a few different ways in which dielectric material can be polarised.

But like I said, I suspect you did not really mean it like this, did you?

Hope it's of some use...

Regards,
Koen

[nul-points]

you should scale this up to the kW range and apply for the OU prize.

as Crocodile Dundee *might* have said:
  "That's not a prize!...."   

[nul-points]

sorry, couldn't resist the flippant response there!

If (and i think this has to be a pretty BIG 'If'!) this phenomenon allows itself to be  translated from a bizarre Laboratory refutation of the understandably skeptical "Ya cannae change the Laws of Physics, Cap'n" point of view, to find a useful place in the home and industry, then i think everyone will agree that this, in itself, is a prize worth having

personally, i suspect the odds of achieving this aren't favourable or someone would have done it by now

the excess energy is manifesting both in the capacitor's stored energy and the charging energy of the associated work done in storing it - it appears that both energies need to be utilised to achieve efficiencies > 100%

the charging energy appears at present to be confined to dissipative material, ie resistive, not inductive or capacitive

this poses an interesting challenge in converting a 'spiky', low-voltage, AC waveform back into something we all know and love - try and do this efficiently, without using an inductor or capacitor (i include transformers under the heading 'inductor')

so, my best bet for now, if we can achieve anything useful at all, is an overunity fan heater!  

all the best
s.