Overunity.com Archives is Temporarily on Read Mode Only!



Free Energy will change the World - Free Energy will stop Climate Change - Free Energy will give us hope
and we will not surrender until free energy will be enabled all over the world, to power planes, cars, ships and trains.
Free energy will help the poor to become independent of needing expensive fuels.
So all in all Free energy will bring far more peace to the world than any other invention has already brought to the world.
Those beautiful words were written by Stefan Hartmann/Owner/Admin at overunity.com
Unfortunately now, Stefan Hartmann is very ill and He needs our help
Stefan wanted that I have all these massive data to get it back online
even being as ill as Stefan is, he transferred all databases and folders
that without his help, this Forum Archives would have never been published here
so, please, as the Webmaster and Creator of these Archives, I am asking that you help him
by making a donation on the Paypal Button above.
You can visit us or register at my main site at:
Overunity Machines Forum



Thane Heins Perepiteia Replications

Started by hartiberlin, May 28, 2009, 05:54:52 PM

Previous topic - Next topic

0 Members and 8 Guests are viewing this topic.

CRANKYpants

Quote from: Kator01 on August 01, 2009, 08:37:15 PM
Mr. T,

you have to include the reactive-power-part because to drive this part needs energy also.

The "total real power" to drive the total "appearant power" must be compared with the real output on the load.

I made a suggestion how to measure this with a diffrerent driver-circuit not using the grid.

So again my question : How are the figures you measure concerning the appearant power-input in the primary ?

Regards

Kator01

HERE IS THE RESPONSE TO YOUR POST USING THE PYTHAGOREAN THEOREM TO DOUBLE CHECK THE NUMBERS - WHICH CAME FROM A DEMO I GAVE TODAY.

T

These are the numbers from today:

Input Voltage = 120.1 V
Input Current = 0.003 A
Scope Observed Power Factor = 0.309 or 72 degrees

True Primary Power = V x I x cos72
                             = 120.1 x 0.003 x .309
                             = 0.111 W

Reactive Primary Power = V x I x sin72
                                   = 120,1 x 0.003 x .951
                                   = 0.343 VAr

Apparent Primary Power = V x I
                                           = 120.1 x 0.003
                                        = 0.36 VA

CHECK

Apparent Primary Power = (square root of) Real Power^2 + Reactive Power^2
                                        = (square root of) 0,111^2 + 0.343^2
                                        = 0.36 VA

broli

Thane some people here would like you to put all kind of shows up. You don't need to be a genius to understand power factors and how you can EASILY get them from a scope. But apparently you have to repeat the same thing 1000 times before it gets through to them. You have shown enough times that you get more out that in. So why don't we all cut the crap and replicate it already.

The main discussion should be about how to get as much people as possible to replicate it. Where to find the cheapest materials to do so by sharing links or ebay offers.

Kator01

Broli,

if you want to adress me, please do so in a direct way. If you think I just want to keep people  busy here with unqualified remarks you should get back to my post #518 on page 52,  read and understand first what I am talking about. Then you should read Thanes first answer #563 on page 57.

Now the point is this :

The mere assumption that driving the primary coil of a transformer with no load at the secondary is for free is simply wrong. It is similar to a free wheeling very low-frequency-oscillator with no extra capacitor but with the coil-capacitance acting instead so  that you have a parallel LC-Tank.
Driving such an LC-Tank in idle-mode needs energy otherwise any LC-tank would just run on its own. If you can present such a circuit here I will be the first to buy it you have my word.
I hope you get this point now.

Driving  an high- or low-frequency LC-tank makes no difference in this regard. The only difference with hf-tanks is that they radiate em-waves which needs still more energy.

In a low-frequency-tank with an iron-core involved you need power to overcome the magnetization-losses even, if you are idle mode.

In order to find the correct number of this minimum-necessary power we need a 50 hz oscillator which is independent of the grid, for example a wien-oscillator with a push-pull-output-stage. Only with such a setup we will be able to measure the real  input. Period

I will not repeat myself.

Regards

Kator01

broli

Quote from: Kator01 on August 06, 2009, 03:09:36 PM
Broli,

if you want to adress me, please do so in a direct way. If you think I just want to keep people  busy here with unqualified remarks you should get back to my post #518 on page 52,  read and understand first what I am talking about. Then you should read Thanes first answer #563 on page 57.

Now the point is this :

The mere assumption that driving the primary coil of a transformer with no load at the secondary is for free is simply wrong. It is similar to a free wheeling very low-frequency-oscillator with no extra capacitor but with the coil-capacitance acting instead so  that you have a parallel LC-Tank.
Driving such an LC-Tank in idle-mode needs energy otherwise any LC-tank would just run on its own. If you can present such a circuit here I will be the first to buy it you have my word.
I hope you get this point now.

Driving  an high- or low-frequency LC-tank makes no difference in this regard. The only difference with hf-tanks is that they radiate em-waves which needs still more energy.

In a low-frequency-tank with an iron-core involved you need power to overcome the magnetization-losses even, if you are idle mode.

In order to find the correct number of this minimum-necessary power we need a 50 hz oscillator which is independent of the grid, for example a wien-oscillator with a push-pull-output-stage. Only with such a setup we will be able to measure the real  input. Period

I will not repeat myself.

Regards

Kator01

Yes, uhuh,I see.... but maybe you have dodged the fact that the output has a resistive load. I will not repeat myself.

CRANKYpants

Quote from: broli on August 06, 2009, 07:54:33 AM
Thane some people here would like you to put all kind of shows up.

I AM AFRAID I DON'T UNDERSTAND THIS STATEMENT?

QuoteYou don't need to be a genius to understand power factors and how you can EASILY get them from a scope. But apparently you have to repeat the same thing 1000 times before it gets through to them.

I HAVE TO REPEAT IT 1000 TIMES TO MYSELF BEFORE I GET IT OR AT LEAST START TO TRUST THAT IT MAY BE REAL...

QuoteYou have shown enough times that you get more out that in. So why don't we all cut the crap and replicate it already.

AS FAR AS I KNOW AT LEAST 4 PEOPLE ARE NOW DOING IT INCLUDING A TRANSFORMER COMPANY AND SOON TWO.

QuoteThe main discussion should be about how to get as much people as possible to replicate it. Where to find the cheapest materials to do so by sharing links or ebay offers.

IN A FEW DAYS WE SHOULD HAVE A FACTORY MADE CORE WITH OR WITHOUT WINDINGS SO ANYBODY THAT WANTS ONE CAN HAVE IT.

T