EEVblog #708 - Free Energy BULLSHIT!

[IanB]

Thank you for the reply...What I was getting at is a light bulb gives off heat and light yet in this circuit there is no difference in the total finishing voltage in all capacitors in the system between a 13 awg wire and the light bulb. How is that possible under conventional engineering laws? That is what I do not understand. People say a light bulb uses energy yet it appears not to in this circuit.

Fat wire: all the lost energy is wasted and thrown away

Light bulb: some of the lost energy is captured and turned into useful light instead of being thrown away

You are trying quite hard to pretend to misunderstand, aren't you?

[parbro]

When we look at the first capacitor, the total charge is not the important thing, the total stored energy is the important thing. The total stored energy is given by a different formula, involving the square of the voltage. The energy, E1, stored in the first capacitor, is this:

    E1 = 0.5 * 10,000 µF * 24 V * 24 V = 2,880,000 µJ = 2.88 J

Here, the "J" means joules, a unit of energy. Energy is what we pay for when we buy electricity from the utility company, only they measure it by kilowatt-hours, or kWh. It's the same thing though.

Let's now see what we have after we join the capacitors together and equalize them. The new energy E2, is this:

    E2 = 2 * 0.5 * 10,000 µF * 12 V * 12 V = 1.44 J

Look what happened. It is only half the energy we started with! We have "lost" half of our valuable energy.

Where did it go? It was wasted, in fact. Lost as heat due to the resistance in the wires.

Given that your equations used did not describe the energy loss in the wires, would it be more accurate to say the 50% loss of energy was the energy required to charge the second capacitor? Or that this energy loss is inherent to the charging of capacitors?

[eneuro]

Here is a patent that covers what we are discussing on this site...
http://www.tuks.nl/pdf/Patents/Benitez/GB191514311A.pdf

I like patents especially those from XVIII/XIX and of course I'm Tesla fun 
Probably made a little bit more complicated circuit than they suggest, but wow, we are in XXI century and for extraordinary claims more advanced technology is needed 

Let me know if you want experiment with this enhanced circuit from this thread and for $10 you can get as many such small (tiny) PCBs as you want-to get more free energy you need of course more such small generators 
12V old car battery can be used there-forget about crappy 1.5V alkaline-we're in renowable energy age 
This was optimized in Spice and oscilator frequency is increased, so it is much more efficient design (150%-178%) can be achived very easy....
However, it needs to be soldered using special kind of soldering technique-It is called SMA (Surface Mount Art) 
If you buy 100 such boards, you will get SMA manual for free 

[max_torque]

3) That circuit still has me a little confused... As the light bulb has the same results as a copper wire does, they both fill the capacitors to 9.3volts respectively. Why is there not more of a loss with a resistive light bulb than a 13awg copper wire you would use in house wiring? That one I do not understand. I will do the tests though to find out. If anyone can answer this please let me know.


-Dave Wing

That is actually quite a good question to ask people learning about electronics!

After all, the start state, and the end state are the same no matter what the resistance of the kink between the capacitors is.

So, why does the circuit "loose" the same amount of energy in either case?  The answer is in the 4th dimension!  And of course is "TIME".   All that the resistance changes in this circuit is HOW LONG it takes for it to achieve equilibrium.  For example, you could expend 10joules worth of energy by expending 10W of power for 1 second, or by expending 1W for 10seconds.  In either case you have "lost" the same amount of energy.

Now the only difficult bit comes when you look at how resistance "looses" power, as it has a square term in the calculation.  So the power in a resistance is IsquaredR (I x I x R).

So when you replace your bulb, with say 10ohms of resistance with a bit of wire with say just 1 ohm of resistance, the equilibrium point occurs 10x faster, with 10x more current average flowing. So, lets say in:

Case 1: with a 10 Ohm bulb in series, it takes 10sec for equilibrium, and 1Amp flows:
The average loss in the bulb (IxIXR) is 1 x 1 x 10 = 10 Watts, and 10watts for 10sec = 100 Joules

Case 2: with a 1 Ohm wire in series, it takes just 1 sec for equilibrium and 10 Amps flow:
The average loss in the bulb is 10 X 10 X 1 = 100 Watts, and 100 Watts for 1 second is, 100 joules



See, it's amazingly elegant how the real physical world works, to the point we don't really need to be making up psuedo science or WooWoo or whatever!!

[IanB]

Thank you for the reply...What I was getting at is a light bulb gives off heat and light yet in this circuit there is no difference in the total finishing voltage in all capacitors in the system between a 13 awg wire and the light bulb. How is that possible under conventional engineering laws? That is what I do not understand. People say a light bulb uses energy yet it appears not to in this circuit.

Case 1: Big fat wire. You just put a short circuit right across your power bank. All the energy got thrown away with a flash and a bang.

Case 2: Bulb. You still drained the power bank, but now some of the energy got changed into light by the bulb.

You need to explain why you think the bulb appears not to use any energy.

[IanB]

Given that your equations used did not describe the energy loss in the wires, would it be more accurate to say the 50% loss of energy was the energy required to charge the second capacitor? Or that this energy loss is inherent to the charging of capacitors?

That's a curiously insightful question, and in fact there is an energy loss inherent in charging capacitors if you just connect them to a power supply.

Let's say we want to charge a 10,000 µF capacitor up to 12 V using a 12 V DC power supply.

When the capacitor has finished charging, the energy stored inside it is given by:

    E1 = 0.5 * 10,000 µF * 12 V * 12 V = 0.72 joules

But the charge in the capacitor that had to be supplied by the power supply is:

   C1 = 10,000 µF * 12 V = 0.12 coulombs

Therefore the energy drained from the power supply is:

   E2 = 0.12 coulombs * 12 V = 1.44 joules

We find that only half the power consumed by the power supply ended up stored in the capacitor. There was 50% of wasted energy!

[DaveWing]

Explain something to me please. If you power a light bulb directly across one 12 volt battery and it draws one amp at 12 volts. Where is the current going? What gets converted into light and heat?

The potential (chemical) energy in your battery. Once all that chemical energy in your battery gets converted into light/heat, then the battery is dead, no more potential energy.

The above is a quote from admin.

If the above quote is true, that potential energy is converted into light/heat how can we pass anything  into the parallel capacitors at all? Yet we see some energy made it through and can be used again... So the system captures some of the energy that flows through the bulb for re-use. The normal configuration, that is directly across a battery or capacitor positive/ negative terminals gives no return for re-use.

I hope you understand my question.

-Dave Wing

[miguelvp]

Explain something to me please. If you power a light bulb directly across one 12 volt battery and it draws one amp at 12 volts. Where is the current going? What gets converted into light and heat?

The potential (chemical) energy in your battery. Once all that chemical energy in your battery gets converted into light/heat, then the battery is dead, no more potential energy.

The above is a quote from admin.

If the above quote is true, that potential energy is converted into light/heat how can we pass anything  into the parallel capacitors at all? Yet we see some energy made it through and can be used again? So the system captures some of the energy that flows through the bulb for re-use. The normal configuration, that is directly across a battery or capacitor positive/ negative terminals gives no return for re-use.

I hope you understand my questions.

-Dave Wing

Because the capacitors increases the load, it would be like placing bulbs with the equivalent resistance of those capacitors.

Without the capacitors no extra load will be demanded from the battery. So adding extra loads just makes your battery to deplete faster and more energy wasted in heat on those capacitors that you are charging because of the internal resistance in the caps.

[IanB]

If the above quote is true, that potential energy is converted into light/heat how can we pass anything  into the parallel capacitors at all? Yet we see some energy made it through and can be used again? So the system captures some of the energy that flows through the bulb for re-use. The normal configuration, that is directly across a battery or capacitor positive/ negative terminals gives no return for re-use.

Nothing was captured for re-use. This isn't what happens.

You start out with some energy. You end up with less energy. This difference is what is lost or consumed. It's your choice whether to use it or lose it.

But there is no return for re-use. Gone is gone.

[Fungus]

Yet we see some energy made it through and can be used again?

Nope.

If any energy "got though" it's because it was never used in the first place. It was shared between two components which are connected in series

Imagine two LEDs in series - you can light both of them up, but...

you need twice the voltage to do it.

The second LED isn't somehow re-using the energy that came out of the first one, it takes extra input voltage to light it up.

But anyway, I'm done here. You're not interested in the truth, any further effort in your direction just makes me the idiot.

[DaveWing]

Yet we see some energy made it through and can be used again?

Nope.

If any energy "got though" it's because it was never used in the first place. It was shared between two components which are connected in series

Imagine two LEDs in series - you can light both of them up

but you need twice the voltage to do it.

The second LED isn't somehow re-using the energy that came out of the first one, it takes extra input voltage to light it up.

But anyway, I'm done here. You're not interested in the truth, any further effort in your direction just makes me the idiot.

Sorry but the flow must go through the load there is no other way.

-Dave Wing

[Galaxyrise]

Given that your equations used did not describe the energy loss in the wires, would it be more accurate to say the 50% loss of energy was the energy required to charge the second capacitor? Or that this energy loss is inherent to the charging of capacitors?

We find that only half the power consumed by the power supply ended up stored in the capacitor. There was 50% of wasted energy!

In the math I did above, the energy was lost to resistance.  The amount of resistance did not dictate the amount of energy lost, only the time it took, so I'm not surprised the energy loss can be computed with R cancelling out.  I think careful inspection of the derivation of those formulas assumes finite current.

[Galaxyrise]

Yet we see some energy made it through and can be used again?

Sorry but the flow must go through the load there is no other way.
-Dave Wing

Current is not energy!

[hammy]

A lot of discussion about basics explained in AoE on page 2, chapter "1.01 Voltage and current".

This troll knows exactly how to push someones buttons. 

[mrflibble]

This troll knows exactly how to push someones buttons. 

No shit. I must say though ...

@DaveWing:
Well played. One of the better played trolls I've seen in a while. Usually it's boringly and sadly transparent. You managed to stay juuust within the envelope of doubt for a pretty long time... but integrated over all messages the mistakes are there. Woopsie, better luck next troll.

No doubt you will come with "but but I am misunderstood, I am just trying to learn". Too late, you overplayed it. Anyways, as other posters have said, even if this is a troll (which it turns out to be), then it's not wasted time. Other people will read the posts as well, and may learn something regarding common mistakes when calculating power. You just played a few people for your amusement, but oh well, the by-product of a good explanation for the benefit of others is worth it.

Well, actually you spoiled it early on already, but subtle enough. Anyways, now scoot. Fun's over. Next forum, next nickname.

[DaveWing]

If the above quote is true, that potential energy is converted into light/heat how can we pass anything  into the parallel capacitors at all? Yet we see some energy made it through and can be used again? So the system captures some of the energy that flows through the bulb for re-use. The normal configuration, that is directly across a battery or capacitor positive/ negative terminals gives no return for re-use.

Nothing was captured for re-use. This isn't what happens.

You start out with some energy. You end up with less energy. This difference is what is lost or consumed. It's your choice whether to use it or lose it.

But there is no return for re-use. Gone is gone.

If you put it to ground you lose it. That is my point. The other point is why lose it? Please explain how does the potential get into the parallel capacitor section... It has to go through the load... That is the only way.

Or if there is another way the potential gets in the parallel caps fill me in on how it gets there.

-Dave Wing

[IanB]

If you put it to ground you lose it. That is my point. The other point is why lose it? Please explain how does the potential get into the parallel capacitor section... It has to go through the load... That is the only way.

Or if there is another way the potential gets in the parallel caps fill me in on how it gets there.

You seem to have answered your own question. Energy gets into the parallel capacitor section through the load. Some gets consumed by the load and some makes it through into the capacitor bank. So what then is your question?

[DanielS]

Sorry but the flow must go through the load there is no other way.

There is another way: if you have too much voltage for your load that there is "energy to recover" (enough voltage headroom to cram a battery or cap in series) then you would be much better off using a DC-DC converter to provide the correct voltage to the load in the first place. This would be far more efficient than using a bunch of batteries or switched capacitors.

[Smokey]

Ok.... Ok... It's pretty dark again.  Faith has been lost. 
At this point wouldn't it be funny if we all just changed our tunes and started agreeing with him.

OMG You're RIGHT!!!  You really have discovered the mythical free energy!!!  Hurry! Go get patents and investors, and tell all your friends!  You're going to be a millionaire!  We are professional engineers and have been doing this our whole lives, but you just opened our eyes with your kitchen table circuit!!!  Why oh why did I stop at two batteries when I could have just added one more and had success!!!!

[wraper]

He's lost for humanity, don't waste your time. And don't dare to invite him in our secret club, he will corrupt it and leak all our secrets  .

[westfw]

Therefore the energy drained from the power supply is:
   E2 = 0.12 coulombs * 12 V = 1.44 joules

I don't think so.  The energy based on charge has a 1/2 factor in the equation as well.

[EEVblog]

So the system captures some of the energy that flows through the bulb for re-use.

You have been told time and time again that current is NOT energy, yet you refuse to acknowledge this fact.
You either have a severe learning disability or, as others have said, you are very expertly trolling us.
I'm am seriously starting to believe the later, in which case, give me a good reason why you shouldn't just be banned?

[IanB]

Therefore the energy drained from the power supply is:
   E2 = 0.12 coulombs * 12 V = 1.44 joules

I don't think so.  The energy based on charge has a 1/2 factor in the equation as well.

One joule is one watt-second, and one coulomb is one amp-second (or put another way, one watt is one joule per second, and one amp is one coulomb per second).

Given this, suppose the power supply delivers 1 amp for 0.12 seconds to charge the capacitor. 1 amp from a 12 volt power supply is a power output of 12 watts.

Therefore the total charge delivered is 1 A x 0.12 s = 0.12 C. The total energy output is 12 W x 0.12 s = 1.44 J.

(0.72 J is stored in the capacitor and 0.72 J is dissipated as heat in the system.)

((Extra capacitance in a system will lead to a worse power factor and will lead to greater I²R losses.))

(((Moral: Do not charge a capacitor from a voltage source. You will lose. Always charge a capacitor from a current source if system efficiency is important to you.)))

[Smokey]

http://bedinissg.com/
Look who the first "Satisfied Customer" is..
He's already invested $27.  Can't back down now!

"Tonight I spent the better part of six or seven hours going over the information supplied in this document and I can honestly say it was very well written, easy to read, very informative and highly enlightening, material... I seriously think it is the most valuable free energy document ever released, as it truly establishes a VERY solid foundation to build upon. Thank you to all who made this possible!

'With the publication of this book, the "debate" about how the machine operates is over.' (John Bedini Nov.2012)

If you are like me and have spent many hours per day learning, experimenting, compiling information from the internet, dvd's, various written works and so forth, but could not quite get the full handle or understanding of exactly how the SSG machine's circuitry worked and or interacted with the other components within the machine, then you cannot miss this one. It is a must have for those who are just starting out and also for those who think they know it all... Best 27 dollars for enjoyment, I ever spent."
- Sincerely, Dave Wing

[EEVblog]

http://bedinissg.com/
Look who the first "Satisfied Customer" is..
He's already invested $27.  Can't back down now!

It seems our friend Dave Wing has been sprouting all this nonsense on other forums as well:
http://www.energyscienceforum.com/search.php?searchid=472885

Pretty clear we are being trolled by someone who will never ever change their mind.