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Originally posted by dereks
Originally posted by TeslaandLyne
In PART ONE I heard 320%.
If power then a battery say 100v and one amp gives out 100v 3.2 amps
which could charge the battery and run the device and have an 1.2 amp
left over
all they would have to do is remove the battery, use a capacitator and feed the output back into the input. If it continues to run then it is overunity. Now why didnt they do that??
Originally posted by Korg Trinity
The system would still need an initial energy source to run.
Originally posted by dereks
Originally posted by Korg Trinity
The system would still need an initial energy source to run.
So simply start with the battery across the capacitor, start the Orbo and charge the capacitor up then disconnect the battery - if it works as claimed it would continue running!
Originally posted by Korg Trinity
You see a capacitor releases energy at intervals depending on the size of the capacitor.
It's by far a simpler and more efficient system to use the battery route.
garbage, just where do you get that from?
An ideal capacitor is characterized by a single constant value, capacitance, which is measured in farads. This is the ratio of the electric charge on each conductor to the potential difference between them. In practice, the dielectric between the plates passes a small amount of leakage current. The conductors and leads introduce an equivalent series resistance and the dielectric has an electric field strength limit resulting in a breakdown voltage. Capacitors are widely used in electronic circuits to block the flow of direct current while allowing alternating current to pass, to filter out interference, to smooth the output of power supplies, and for many other purposes. They are used in resonant circuits in radio frequency equipment to select particular frequencies from a signal with many frequencies.
Because the battery simply powers the Orbo until it drains, and in the case of the Orbo that takes 7 days.
Originally posted by PrisonerOfSociety
reply to post by Korg Trinity
Sean, is that you?
Originally posted by Korg Trinity
A capacitor only stores energy until a predefined value is met before releasing.
The system would still need an initial energy source to run.
Originally posted by PrisonerOfSociety
reply to post by Korg Trinity
Sean, is that you?
Originally posted by buddhasystem
Originally posted by Korg Trinity
A capacitor only stores energy until a predefined value is met before releasing.
This is pure, unadulterated nonsense. This comes from a person who tinkered with electronics since age 7 and completed formal courses in electronics in college.
The system would still need an initial energy source to run.
Of course. You give it some energy and then disconnect the initial source, just like a battery in your car.
When the switch is closed, current from the battery flows through the circuit, charging the capacitor. When the capacitor is completely charged, it is like a closed tank which is completely filled up, and no further current flows. At that time, the voltage across the capacitor would be equal to the supply voltage of the battery.
Voltage across the capacitor advances from zero (fully discharge) to the supply voltage along some predetermined path with respect to time. If the resistor is small, current flows easily and the capacitor is charged more quickly. If there is a very large resistor, the charging process follows a different path and will take longer to complete. The behavior of voltage versus time is also influenced by the size of the capacitor. If the capacitor’s capacitance is very large, it will require more total energy to fill (the tank is large in diameter), and current flowing through the resistor will require a longer time to charge it.
Originally posted by seethelight
reply to post by Korg Trinity
but if the system was overunity, why would it ever stop?
Over-unity does not mean perpetual motion. It simply means more power out than in.
Originally posted by seethelight
reply to post by Korg Trinity
hmm.... so you're just trying to figure out how effeciently it could use the single charge...
and yeah I was just using short hand with the forever question...I meant energy wise
Originally posted by PrisonerOfSociety
reply to post by Korg Trinity
Over-unity does not mean perpetual motion. It simply means more power out than in.
With all due respect, that's exactly what OU means; when System.Out > Sytem.In then there's a feedback loop which 'perpetuates' energy.
Place an LED on the surplus emf, and it should light without impacting the overall system...isn't that the point of all this stuff?
Originally posted by Korg Trinity
I am tempted to use my credit card and take a day or two off work to go to ireland and get them to do this test.