EEVblog #708 - Free Energy BULLSHIT!

[Muttley Snickers]

every individual customer is billed based only on their usage and supply service charge.

Built into their cost of doing business is "stock shrinkage" which includes unexplained loads (a single private home is probably "down in the  noise"), but even greater (and more constant) is simply losses through the transmission system between the source and all the loads.

I fully understand stock shrinkage, it is a part of what I do.

From what I heard, this non billing of hundreds of customers was an oversight by a number of electricity providers, and even early in the peace we notified them on many occasions that we hadn't received an account, nothing ever got attended to, so we gave up trying.

We were expecting a bill for around eight grand and would have happily paid it, but it turned out that legally they couldnot request any more than 9 months worth prior. I still feel really guilty to this day.   

All good now though.   

[Fungus]

You couldn't specify *just* a voltage, or you could hook up a 30A supply and light up a whole bank of LEDs.

We were talking about an AA battery.

Battery datasheets will tell you the discharge curve/expected lifetime at different current drains.

[DaveWing]

Ok according to Kirchhoff's current law what is going on in the circuit I posted below. I am asking an honest question here.
If you were to put the light bulb across the negatives of 24 volt and a 12 volt battery system? What results can we expect to find?

You really should be able to tell by looking at the picture what will happen, you shouldn't need anyone to explain it to you, even if you don't understand all the details. You can also do the experiment yourself to find out, maybe by using small rechargeable batteries instead of big 12 V ones.

Assuming batteries 1 and 2 start out fully charged, and battery 3 starts out empty, and assuming the blue circle is a bulb, then:

Initially the bulb will light brightly. Over time the bulb will grow dimmer and dimmer until the combined voltage of batteries 1 and 2 equals the voltage of battery 3. At that time the bulb will go out and the system will stop changing.

When the system has stopped changing, battery 3 will contain some of the energy that was previously contained in batteries 1 and 2. The rest of the energy that was contained in batteries 1 and 2 will be lost to the environment, either as light or heat. Some of the heat energy will be lost by the bulb, and some of the heat energy will be lost from the batteries due to their internal resistance and other inefficiencies.

If you do a total accounting of the energy present in the system at the start and the energy present in the system at the end, the energy at the end will be less than the energy at the start by the amount of energy lost to the environment.

So how much battery capacity (a rough guess will suffice) will battery 3 receive from batteries 1&2, provided they all have the same amp hour capacities when fully charged?

So if battery 3 does receive a percentage, in the form of actual battery capacity from batteries 1&2 is that not better than the conventional model that directly loses all and recaptures none of the battery capacity it uses?

What would happen if you connect a capacitor in place of the battery number 3? Could you not power the bulb for a certain time period or until the capacitor potential equalizes with the two batteries 1&2. When the capacitor equalizes disconnect the cap from the circuit and power the light again until the cap is discharged. Repeat process over and over. Yes there will be losses but not much, you input one part energy and used very close to two parts energy.

I will post a video showing the losses in the system, just like Dave has requested, shortly.

-Dave Wing

[Richard Crowley]

You couldn't specify *just* a voltage, or you could hook up a 30A supply and light up a whole bank of LEDs.

Right.  Which is why I said:  "power (voltage, current)"

[Fungus]

So if battery 3 does receive a percentage, in the form of actual battery capacity from batteries 1&2 is that not better than the conventional model that directly loses all and recaptures none of the battery capacity it uses?

Well....yes, but you had to use TWO batteries (1+2) to do that.

If you take out battery 3 then you can also take out battery 2 and still light the lamp. Overall this will be more efficient because the initial charging of battery 2 is lossy and so is the charging of battery 3 when the circuit is running.

What would happen if you connect a capacitor in place of the battery number 3? Could you not power the bulb for a certain time period or until the capacitor potential equalizes with the two batteries 1&2. When the capacitor equalizes disconnect the cap from the circuit and power the light again until the cap is discharged. Repeat process over and over. Yes there will be losses but not much, you input one part energy and used very close to two parts energy.

The average light output will be halved because the voltage across it is constantly ramping up/down between 0 and 24V.

ie. There's no real-world gain.

[Galaxyrise]

What would happen if you connect a capacitor in place of the battery number 3? Could you not power the bulb for a certain time period or until the capacitor potential equalizes with the two batteries 1&2. When the capacitor equalizes disconnect the cap from the circuit and power the light again until the cap is discharged. Repeat process over and over. Yes there will be losses but not much, you input one part energy and used very close to two parts energy.

First, I'm going to use a resistor as a crude approximation for a light bulb.  If I understand your hypothesis correctly, the exact load doesn't matter anyway.  So we have this circuit:


When power is applied, the capacitor is fully discharged so R1 gets the full voltage applied to it, and the maximum current begins to flow.  But this current begins charging the capacitor, and the capacitor voltage begins to rise.  This rising voltage means there's less voltage across the resistor (the sum of the voltages must be the battery voltage), so the current drops.  Eventually the capacitor is charged up. 


We can already see something you I'm not sure you were accounting for: the "bulb" isn't going to be at full brightness very much (the yellow line.)  But now let's do the analysis that IanB talked about: energy in and energy out.  The question here is how much energy did that take from the battery, and how much energy was used by the bulb?  Since current was changing constantly over time, the math is a little non-trivial.  Multiplying the current curve by the voltage gives you power, and then integrate that to get energy.  This is NOT just a current analysis!

We'll treat battery voltage as constant for now and ignore the other "minimal losses" you talked about.  There is a well-known formula for capacitor charging, so I can do this quite precisely.  With the values I used in the above schematic, the battery supplies 1.44mJ in the time it took to charge the capacitor.  How about output? How much energy ends up in the capacitor? This works out to 0.72mJ.  Not surprisingly, doing the calculus to figure out how much energy was dissipated by the resistor yields 0.72mJ.  So it all adds up: half the energy from the battery went to the "light bulb" and half to the capacitor.  Now you propose dumping the capacitor energy into the light bulb, which is fine.  The energy still came from the battery initially.  Energy in = Energy out.

That's not to say the circuit doesn't have interesting properties.  Pulsing the load like this is a form of switching dc to dc converter, and there are situations where they are quite useful.

[DaveWing]

What would happen if you connect a capacitor in place of the battery number 3? Could you not power the bulb for a certain time period or until the capacitor potential equalizes with the two batteries 1&2. When the capacitor equalizes disconnect the cap from the circuit and power the light again until the cap is discharged. Repeat process over and over. Yes there will be losses but not much, you input one part energy and used very close to two parts energy.

First, I'm going to use a resistor as a crude approximation for a light bulb.  If I understand your hypothesis correctly, the exact load doesn't matter anyway.  So we have this circuit:


When power is applied, the capacitor is fully discharged so R1 gets the full voltage applied to it, and the maximum current begins to flow.  But this current begins charging the capacitor, and the capacitor voltage begins to rise.  This rising voltage means there's less voltage across the resistor (the sum of the voltages must be the battery voltage), so the current drops.  Eventually the capacitor is charged up. 


We can already see something you I'm not sure you were accounting for: the "bulb" isn't going to be at full brightness very much (the yellow line.)  But now let's do the analysis that IanB talked about: energy in and energy out.  The question here is how much energy did that take from the battery, and how much energy was used by the bulb?  Since current was changing constantly over time, the math is a little non-trivial.  Multiplying the current curve by the voltage gives you power, and then integrate that to get energy.  This is NOT just a current analysis!

We'll treat battery voltage as constant for now and ignore the other "minimal losses" you talked about.  There is a well-known formula for capacitor charging, so I can do this quite precisely.  With the values I used in the above schematic, the battery supplies 1.44mJ in the time it took to charge the capacitor.  How about output? How much energy ends up in the capacitor? This works out to 0.72mJ.  Not surprisingly, doing the calculus to figure out how much energy was dissipated by the resistor yields 0.72mJ.  So it all adds up: half the energy from the battery went to the "light bulb" and half to the capacitor.  Now you propose dumping the capacitor energy into the light bulb, which is fine.  The energy still came from the battery initially.  Energy in = Energy out.

That's not to say the circuit doesn't have interesting properties.  Pulsing the load like this is a form of switching dc to dc converter, and there are situations where they are quite useful.

Resistive load selection is important, the less the internal resistance the better and more efficient the energy transfer. The diagram as seen below and the video http://youtu.be/X4dJdEDwjgU I posted was not meant to be run as a continuous closed loop system... It was meant to be a pulsed based open loop system. I only ran it closed loop to demonstrate that recovery is very possible, also I wanted to raise the question why we are not doing it with our large powered systems that we use everyday in our daily lives.

Thank you for the reply.

-Dave Wing

[edy]

Another great video Dave! I thoroughly enjoyed the explanations and bringing things back to science! Too often the uninformed public without any science and particularly physics or engineering education are misled by the media, by people who wish to profit, and sometimes unintentional "believers" who don't understand what is going on and resort to mystical explanations rather than picking up a good book on fundamental theories.

One thing I did NOT do is click on the original free energy video that you linked because the last thing I wanted to do is to increase their YouTube video hits, which I assume are monetized. By watching the original video they would just gain more Adsense profits.

Anyways, excellent video as always!

[Galaxyrise]

Resistive load selection is important, the less the internal resistance the better and more efficient the energy transfer.

Well, let's try it with a 1000 ohm resistor.  The energy that ends up in the capacitor will still be 0.72mJ because that's how much energy the 10u capacitor can store at 12V.  And the energy drawn from the battery... calculus... 1.44mJ.  Same as with the 10 ohm resistor!  The difference is that it takes much longer.

[MBY]

Repeat process over and over. Yes there will be losses but not much, you input one part energy and used very close to two parts energy.

I will post a video showing the losses in the system, just like Dave has requested, shortly.

-Dave Wing

But no!!!
To be ignorant is one thing, we are all here to learn something. But you maintain your delusion even when we point it out to you that you will not gain anything, and that's worrisome. You have mixed up the notions of current, voltage, power and energy and this is nothing that's is open for discussion. You are simply wrong.

If you by "part of energy" mean the energy consumed by the bulb when charging the capacitor, you have used two such "parts". One for the bulb, one for the charging. That the bulb is in series with the caps doesn't come in to the equation. So, assuming no losses whatsoever, charing a cap and use the cap to drive a bulb is just a very stupid and convoluted way of drive the bulb.

You cannot save an attojoule with your scheme. You clearly need to go back to basics and learn Ohms law, the power law, the relationship between power and energy and realize that a conductor transmitting electrical energy is not the same thing as using this energy. And, you cannot use Krichhoff's current law locally to measure energy of a whole system globally.

You sir are dead wrong. Totally wrong. There is no discussion at all. Your scheme isn't even reminiscent of real power saving ideas like reusing braking energy. There is nothing, absolutely nothing in your idea that will save, conserve or make clever use of energy. The only thing your scheme does, is maximizing losses. Period.

[DaveWing]

If you do a total accounting of the energy present in the system at the start and the energy present in the system at the end, the energy at the end will be less than the energy at the start by the amount of energy lost to the environment.

And to do that you could discharge the batteries completely after the test and log the results to find out how much energy was left in them. And along with knowing how much energy was in the original batteries to begin with, and also the log of the load power over time, you'll get the result as IanB states.
For a ballpark measurement you could likely ignore the battery and charging losses.
DaveWing should do this before talking any more about this subject, it's just getting painful.

Hi Admin,

Sorry for the pain you receive on my account.

I tried to make a video as per your request.
In any event I hope that this video will help me to understand some of my misconceptions. So here it is please offer up any comments all are welcome to put in their two cents... As they have so done before.


Here is two diagrams one of a conventional system and the other is... After watching the video... Well you decide.

Regarding the video please excuse my rough delivery. Thanks for watching... http://youtu.be/K6CpDaksxh0



-Dave Wing

[Fungus]

Resistive load selection is important, the less the internal resistance the better and more efficient the energy transfer. The diagram and the video I posted was not meant to be run as a continuous closed loop system... It was meant to be a pulsed based open loop system. I only ran it closed loop to demonstrate that recovery is very possible, also I wanted to raise the question why we are not doing it with our large powered systems that we use everyday in our daily lives.

Waffle. Zero information content.

Thank you for the reply.

What conclusions did you draw from it?

(If you're still raising the question "why we are not doing it with our large powered systems that we use everyday in our daily lives" then I guess 'not much'... )

[Trax]

@Trax

You have proposed many amazing technologies but I proposed just one.
Not that I don't like a good sci-fi setup.
100K years is just too much, any technology you put in your space ship would be obsolete in 20 years.
Look at Rosetta spacecraft, it has a camera of 4 megapixels because 10 years ago they couldn't do better.
And also there are 100 billion other galaxies.

Well what I am proposing is technology that is realizable in accord with the laws of physics, what you are proposing is to the best of the words knowledge impossible.

It is irrelevant if the technology will be out dated in 100k years if its done right at the day of manufacturing it will be still working in 100k years and thats all that matters.

Also you ignored the fact that for leaving out solar system with FTL speeds we don't need infinite power but we need some loop whole drive that would bypass the limitations set forth by the theory of relativity, ...

Over unity is for all intents and purposes a nice to have but by no means necessary for us to leave the solar system.

Your argumentation is flawed in any conceivable way sorry...

[edy]

The other HUGE problem with all these "free energy" ideas is that nobody actually understands what "free energy" really means. For example, how many times does someone claim they are gaining "energy" from *nothing* when it is actually just converting something else that is not obviously apparent.

For example, even in this example of the LEDs... what "energy" is actually being created? A bunch of dim red photos popping out of 23 LEDs as a result of work done by a AA battery with a comparatively large amount of chemical potential energy still stored in it. I still see a chemical reaction losing potential while photons are gained. Overall net loss anyways due to the inefficiency of the system, heat loss, etc.

Even in cases where energy appears to be freed from nothing, if you look closer at the system you may find perhaps it is a nuclear reaction. No matter where you look, the conservation laws of physics are still there. No exception. Even if you could blame it on some "quantum" gobbledegook winkidinky, it is still coming from there... it is not "FREE" because conservation must always be observed. Nothing in life is FREE.

[jancumps]

If you do a total accounting of the energy present in the system at the start and the energy present in the system at the end, the energy at the end will be less than the energy at the start by the amount of energy lost to the environment.

And to do that you could discharge the batteries completely after the test and log the results to find out how much energy was left in them. And along with knowing how much energy was in the original batteries to begin with, and also the log of the load power over time, you'll get the result as IanB states.
For a ballpark measurement you could likely ignore the battery and charging losses.
DaveWing should do this before talking any more about this subject, it's just getting painful.

Hi Admin,

Sorry for the pain you receive on my account.

I tried to make a video as per your request.
In any event I hope that this video will help me to understand some of my misconceptions. So here it is please offer up any comments all are welcome to put in their two cents... As they have so done before.


Here is two diagrams one of a conventional system and the other is... After watching the video... Well you decide.

Regarding the video please excuse my rough delivery. Thanks for watching... http://youtu.be/K6CpDaksxh0



-Dave Wing

I'm not going to comment on the schema and theories, but I'd like to give you two off-topic  'A's:
1 for guts and bravery: kudos on you to stand up and work along in this conversation
1 for the video and voice quality : it was easy to watch and listen to.

[GlowGale]

Hi Dave,

Your video had me laughing so much I had to come to the forum and say hi!

Man, his original posts are real troll nuggets. Throwing around unrelated terminology & using fancy terms to disguise his lack of circuit knowledge. The funny thing is he could have used one toroid with 3 windings and it would work better with less space than a stack of 5 toroids. Although I heard quantum vacuum energy sticks better to black tape...

If you have a metric ton of cash laying around, the most hardcore way to debunk most of these circuits would probably be to measure the outbound energy flux (mostly heat & light, also electrical parasitics?) and integrate it over the battery lifespan. That amount of shed energy would be equal to the chemical energy capacity of the AA-cell and everything's dandy according to the law of energy conservation.

Of course, the electrical way to debunk is a lot more interesting. Many thanks for the big refresher on how these boost converters work!

Greetings,
GlowGale

[Fungus]

I tried to make a video as per your request.
In any event I hope that this video will help me to understand some of my misconceptions. So here it is please offer up any comments all are welcome to put in their two cents... As they have so done before.

Here is two diagrams one of a conventional system and the other is... After watching the video... Well you decide.

I see one immediate (and major) flaw: Capacitor charging is not linear. Adding together the three voltages at the end does not tell you that the total energy in the system is the same.

Energy in a capacitor is proportional to the square of the voltage.

24.9 * 24.9 = 620
(7.92 * 7.92) * 3 = 188

So the total energy in the system went down by about 70% when you lit up the bulb (from 620 to 188)

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html

Regarding the video please excuse my rough delivery. Thanks for watching... http://youtu.be/K6CpDaksxh0
-Dave Wing

Delivery, camera, lighting, etc. was perfect:  10/10

[morrone]

Here is two diagrams one of a conventional system and the other is... After watching the video... Well you decide.

Before we can explain more complicated circuits to you, you are going to have to take some time to understand the basics.  Consider this circuit:



There will be 1.2A flowing in this circuit.  The same current flows through both resistors.  It does not change since these parts are in series.  If the two resistors were wired in parallel, then the current would be split between them.  But they are not wired in parallel; they are wired in series.

While the current remains the same, the voltage drop over each in-series component will be different.  And the sum of the voltage drops across the in-series components will be equal to the total voltage supplied by the battery.

In this example, since the resistors have exactly the same value, exactly half of the voltage will drop across R1 and the other half of the voltage will drop across R2.  The battery is supplying 24V, so the voltage drop across each of the resistors is 12V.

We can now calculate the power used by each resistor.   Power equals current times voltage.  12V * 1.2A = 14.4 W

In other words 14.4 Watts are used by R1, and another 14.4 Watts are used by R2.  The total power used in the circuit is 28.8 Watts.

Notice that R2 does not reuse the power used by R1.  Some power is spent in each, and all of the power in total comes from the battery.

If you now look at your simpler circuit with three batteries and a light bulb, you will see that it really isn't all that different from my example.  You have two 12V batteries in series, whereas I made up a single 24V battery in my example.

Instead of two resistors in series like my example, you have a light bulb and a third battery wired in backwards.

A light bulb and a resistor are nearly identical from a circuit analysis perspective.  The light bulb turn power into light and heat, the resistor turns all of the power only into heat.  But they both use up power, and it isn't really important where that power goes, it is gone.

Granted a battery is not quite as simple as a resistor, but the division of power between the light bulb and the battery works in much the same way as the division of power between the two resistors in my example.

The important thing to remember is that with components linked in series, the voltage drop will be split between them.  The split will not always be even, because the ratio of the voltage split is determined by the ratio of the resistances of the components.  But it _will_ be split.  Therefore the power expended in the circuit will also be split amongst the components.

No magic.  No reuse or recycling of power.

[Howardlong]

If you do a total accounting of the energy present in the system at the start and the energy present in the system at the end, the energy at the end will be less than the energy at the start by the amount of energy lost to the environment.

And to do that you could discharge the batteries completely after the test and log the results to find out how much energy was left in them. And along with knowing how much energy was in the original batteries to begin with, and also the log of the load power over time, you'll get the result as IanB states.
For a ballpark measurement you could likely ignore the battery and charging losses.
DaveWing should do this before talking any more about this subject, it's just getting painful.

Hi Admin,

Sorry for the pain you receive on my account.

I tried to make a video as per your request.
In any event I hope that this video will help me to understand some of my misconceptions. So here it is please offer up any comments all are welcome to put in their two cents... As they have so done before.


Here is two diagrams one of a conventional system and the other is... After watching the video... Well you decide.

Regarding the video please excuse my rough delivery. Thanks for watching... http://youtu.be/K6CpDaksxh0



-Dave Wing

What you didn't show, but rather conveniently assumed, is what happens when you actually do put those caps in series and re-inject the circuit. Sure, you get 24V, but putting those three caps in series might give you 24V but the capacitance is now only 3,333uF.

You need two equations, one for the energy held in a capacitor given its capacitance and the voltage across its terminals, and one for calculating the capacitance of several caps in series.

Your setup loses 2/3rds of its energy mostly in the light bulb.

See here:

[edy]

Yes, I was thinking about that as well. It is like a pendulum swinging... and each time it passes through the lowest point it drives a "wheel" that spins for a bit. Eventually the pendulum runs out of energy. As the pendulum is raised, it has "potential" (like the capacitor plates charged to a certain voltage) and as it is allowed to swing it builds up kinetic energy (like the current travelling through the light bulb it does work), but then it does not raise back up to the same potential energy. In this case the calculations show 2/3rds loss each pass or swing.... Energy being lost as light and heat by the bulb. Like a super-bouncing ball or pendulum.

[snoopy]

Hi All,

This is my first post here, after signing up.

I do believe in good science and proper evaluation and I thank Dave for the time he spent on the debunking video. So in light of this... I will ask have any of you heard of John Bedini and his SG Enegizer? Dave mentioned the Joule Thief so I suspect he may of heard of John's SG circuit.

-Dave Wing

JOHN BEDINI UNVEILS 14FT. HIGH MONOPOLE MOTOR AT CONFERENCE


 

Why do all of these silly "free energy motors" use batteries ? Why not replace the batteries with a suitable capacitor with an initial charge. If the contraption is unity or over unity as claimed then the charge on the capacitor will never decrease or increase if it is over unity. I'm almost certain that the capacitor will eventually lose all of its charge ?

[jancumps]

JOHN BEDINI UNVEILS 14FT. HIGH MONOPOLE MOTOR AT CONFERENCE

Whow, loads of capitals. Shift key defect?

[DanielS]

Resistive load selection is important, the less the internal resistance the better and more efficient the energy transfer. The diagram as seen below and the video http://youtu.be/X4dJdEDwjgU I posted was not meant to be run as a continuous closed loop system... It was meant to be a pulsed based open loop system. I only ran it closed loop to demonstrate that recovery is very possible, also I wanted to raise the question why we are not doing it with our large powered systems that we use everyday in our daily lives.

You do not see it used simply because it does not work that way: if you use two batteries and a shunt element to limit current into a third battery, half the power gets used by the shunt and the other half goes into the third battery, minus the batteries' internal losses. After the first iteration, you now have two dead batteries and one (mostly) charged one assuming it started discharged - if the third battery started charged, then you wasted half of your first two batteries as heat into the third one since batteries dissipate excess charge as heat - at least they they do not vent gas or explode.

Now you can take your third battery and run your 12V load for about the same duration as you did with the first two. By the time the third battery is depleted, you will have run your load about twice as long as a single battery could have, which you could have achieved by simply connecting batteries #1 and #2 in parallel, and omitting battery #3 altogether. You save the expense, weight and space of a whole battery.

Battery #3 does not contribute anything useful to your circuit; all it does is compensate for ridiculously poor initial design in an expensive (if you are going to run a 12V load, run it off 12V), potentially dangerous (overcharge hazards) and impractical (need to swap batteries around) way. Put two batteries in parallel, ditch the third and call it a day: simpler, safer, no battery handling required, no additional efficiency loss from having to charge the unnecessary third battery.

[snoopy]

JOHN BEDINI UNVEILS 14FT. HIGH MONOPOLE MOTOR AT CONFERENCE

Whow, loads of capitals. Shift key defect?

Not to mention all of the true believers at the demo

[IanB]

Resistive load selection is important, the less the internal resistance the better and more efficient the energy transfer.

Are you saying you want the load to have a low internal resistance? In your diagram you show the blue circle as a "Low Resistance Load".

Now, this is where you run into trouble. Let's take this to the limit and assume we can make a "load" with zero resistance. Perfect, right? Except, no, not at all. If a load has no resistance it will consume no power and do no work. It will be completely useless. In order for a load to be useful it must have resistance. There is actually a "best" resistance to choose, not too small and not too large. At either extreme, low resistance or high resistance, the power will approach zero and nothing useful will be gained. Somewhere in the middle, with the right resistance, power transfer is maximized.

Do you ever go to the gym and work out? If you want a good workout you have to set the resistance on the machines, right? If you set the resistance to zero, how much of a workout do you think you will get? Setting the resistance to zero will do you no good. It will be like doing weight training with a feather.