COP 17 Heater | Rosemary Ainslie

[witsend]

Was waiting up for gotoluc's data. Can't keep the eyelids open. I'll check this in the morning.

'night folks.

[Aaron]

@mh

When the coil has current going through it it is acting like a compressed spring. When the MOSFET switches off, it is like you have instantly taken the pressure off of a compressed spring. The spring goes "Boing!" at a high velocity, just like the coil goes "Boing!" at high voltage.

That was your Zen moment.

1. Caps will smooth the battery activity, however, for this circuit, we need to keep caps out of the main circuit and off the battery.

2. You're telling me what to test from behind a computer, while I'm at the bench actually gaining empirical evidence of what I see that refutes your ideas.

3. Yes, the coil is like a spring getting compressed, where the compression generates a LOT of heat during that process. The heat is like work expended to lift a ball to a certain height. The coil getting charged is an object being lifted to a certain height. Turning off the mosfet is like releasing the ball so that free radiant potential kicks in a collapses the magnetic field into a spike or pushes the ball to the ground, to bounce and bounce and bounce where all the upwards lifts on the bouncing ball is more than it took to lift it.

You may debate this, but but your viewpoint won't change reality. You may do better by actually allowing yourself to simply see nature as it already is.

[gotoluc]

Hi everyone,

Test stopped Friday 24th at 9pm and measured after 2 hours to allow stabilization.

Supply Battery voltage 12.89

Charge Battery voltage 12.91

Well, I don't quite know what to make of this

Three days ago the battery that first started as supply was at 12.95 and dropped to 12.92 after 24 hours of supplying. It was swapped positions and it charged back up to 13.01 and again swapped position and now dropped down to 12.89

The battery that first was under charge started at 12.82 and went up to 12.90 after 24 hours of charging. It was swapped positions and went down to 12.81 and again swapped position and is now at 12.91

To me, I would conclude that even though the battery under charge are showing voltage gains after a 24 hour charge they seem to loose it (and more) after being swapped back as supply battery. The charge seems more superficial and drops faster after being put to work.

If someone can enlighten me on something that can do to improve the quality of the charge during the charging period so the battery perform better under load (like when charged with amps) it could help the results.

Other than that I see no real gain using this method. We need to keep in mind that the tests are only using 12vdc and could have a different effect using higher voltage.

I will recharge the batteries and try it with 3 identical batteries in series (36vdc) as supply and on the charge side as I have a total of 6 of these 12vdc 5AH sealed lead acid batteries on hand.

I'm open to suggestions.

Luc

[MileHigh]

Hi Aaron:

Just for fun, the spring. I will gladly give you an example from nature.

In nature, you see energy manifest itself as the "through variable" and the "across variable." A compressed spring, force travels through the spring, the "through variable". And there is a velocity between the two ends of the spring, the "across variable." The through variable times the across variable gives you power, force x velocity.

For electrical energy the through variable is current and the across variable is voltage. So if you follow the analogy a charged inductor has current going through it, and no voltage across it's terminals. The voltage times the curent gives you power.

Without going into all of the details, any way the inductor reacts is the same as the way the mechanical spring acts, when you look at the through and the across variables for both systems. That also applies for resistors and capacitors.

Here is a simple mechanical model for what this circuit does: You lay a HO scale train track across your bench and you have a boxcar that you are going to push back and forth down the track. You need to create some friction to slow the boxcar down like it is being pushed through mud. It's a thought experiment so it's thick chocolate pudding ooze. On one end of the boxcar you attach a moderately stiff spring that you can push on with your finger.

Then you push on the spring at a fixed speed and watch the boxcar move through the thick ooze of the choclolate pudding. You notice the faster you push the more the spring compresses. You push for exactly three seconds at a constant velocity and then stop cold. When you stop, you notice the boxcar creep forward in the ooze as the spring decompresses against your stationary finger. Then after 5 seconds you push again. You notice the spring compressing before the boxcar starts moving, and then for a few seconds you are "plowing through the ooze" and pushing the boxcar forward at a constant speed, you stop cold and see the boxcar inch a bit further forward through the ooze as the spring decompresses again.

It doesn't matter how thick the ooze is, you have a magic finger that can push forward at the same speed no matter how much resistance you feel.

Now, can you connect the dots with respect to that analogy?

When your finger moves forward at a constant speed, that's when the MOSFET switches on and you are providing voltage to the circuit (voltage = velocity). When your finger stops moving, the MOSFET switches off.

The chocolate pudding is the resistive part of the coil-resistor. When you move through the thick ooze you are exerting a lot of power, and that power is burned off as heat in the ooze, just like bending a coathanger.

When you push the boxcar through the ooze, you are turning the mechanical power you are providing, force x velocity (your finger) into heat. You push your way through the ooze and the ooze heats up. i.e.; the resistor heats up.

When you stop, the spring starts to decompress. Very important point here: When you stopped applying force into the spring by stopping cold, the spring now starts to decompress and keeps pushing on the boxcar with exactly the same force that you were originally pushing with. Can you see that? The force pushing the boxcar forward remains the same even though you stopped pushing. Just like an inductor will keep the same current flowing.

And think about when you stop, and notice the boxcar move forward from the spring decompression. You create a litle bit of hot ooze when that happens. That's the coil-resistor discharging through itself and the diode after the MOSFET switches off.

Now the short version: You start pushing, the spring stores some of your expended energy before the boxcar is plowing through the ooze and burning off the power you are supplying with your moving finger and heating the ooze. When you stop, the energy you put in to compress the spring is the same ernergy that you see when the spring decompresses and pushes the boxcar forward a bit through the ooze. Wait five seconds and repeat.

A train set in mud with springs, with a "flown in" magic velocity finger that starts and stops. That's a very accurate model for this circuit.

MileHigh

[poynt99]

That's a great analogy for the high-current low voltage kickback mode MH
For fun...

How about the low-current high voltage kickback mode (i.e. no diode present)? This might do:

Right at the moment the pushing finger stops dead cold, it is replaced with a very light and small grain of sand, and in fact it gets glued to the end of the spring.

Now instead of the boxcar edging forward (in the same direction) a little more just as it did before, the force/friction balance is inverted and the boxcar kicks ahead only a small fraction of a fraction of what it did before. However, the spring with the grain of sand glued to the end gets kicked at a very high velocity back in the opposite direction due to the decompressing spring! And it will oscillate back and forth a little as the spring settles down.

.99

[MileHigh]

.99:

Thanks for the compliment! How about this:

The mud is hard to imagine, so it's easier to start from the electrical side. Thinh of a high value resistor - lots of voltage with very little current going through it. Translate that to the train set - for very high velocity, you only need very little force. In other words, the spring is set to fly!

So if you make the resistance much higher you expect to get faster kick-back from your spring. Therefore, higher resistance = thinner mud.

So when the coil is has a lot of current going through it and a 1 Mohm resistor across it, you have a big compressed spring connected to the boxcar, and there is almost no mud amymore and the mud is very thin.

Now the boxcar is actually a magic boxcar. It has almost no mass. So when the MOSFET switches off you have a nearly massless compressed spring and a nearly massless boxcar, and the open highway ahead.

The spring is released and the nearly massless boxcar takes off like a bullet at extremenly high velocity through the 1 Mohm mud, which is more like fog than mud.

There you go!

MileHigh

[witsend]

gotoluc - I'll wait to see when you're on line to post a reply. I'll be back here in about 4/5 hours. Check the time and do the math. I'd be glad to find out a few things.

May I add my endorsement to Armagdn03's opinion of you and your work. In fact I think it's shared everywhere. You are unquestionably my role model. And you have been remarkably tolerant and remarkably patient.

Many, many thanks for everything.

Rosemary

EDIT - If you can't make it then I'll wait patiently. Luc I have to duck out for a couple of hours. It's now our time 14.45

[Aaron]

non-equilibrium thermodynamics

When you stop, the energy you put in to compress the spring is the same ernergy that you see when the spring decompresses and pushes the boxcar forward a bit through the ooze.

I know these are details that don't have to do with replication of the circuit as far as building and sharing. But until I'm ready to show power measurements, I like to discuss these concepts until then.

MH, I admit I love vivid analogies because they're easy for everyone to learn from, but this is a point I want to make.

It is no different than lifting a bouncy ball. It takes 1 joule of work to lift a ball (that weighs about what a small apple would weigh) to 20 cm high.

After lifting that ball against gravity (compressing the spring) to 20cm, that 1 joule of work is work expended never to be seen again. We never stored any gravitational potential in the ball since there is no "storing" potential - we can only create a potential difference to have potential to move from one point to the other and that is what is called energy. Our 1 joule of work was 100% dissipated and this is why it requires 1 joule of work to lift it that high.

During the course of the WORK of the mechanical lifting to 20cm. We got out of it what we put in at that point. We got 20cm of lift from 1joule that we'll never see again. It was dissipated in the resistance against gravity.

Again, the concept of we get out what we put in is this. We put in 1 joule of work and what we get out of it is 20 cm of lift. That lift IS what we get out of it.

But what we have now is a brand new gradient or potential difference established between the ball at 20cm high and the ground. When released, there is no stored energy coming back. That separation in potential differences between ground and the apple at 20cm high is a new dipole, which allows gravitational potential, which is flowing down in my opinion, to be accessed by the apple only when it is in free fall. While still at 20cm, there is no storing anything. There is no intrinsic change in the properties of that apple. The gravity potential flowing thru it is the same as if it was on the ground except the higher up and away from mass (ground), there is actually less gravitational potential available to it since it is in less dense gravity. anyway... a small tangent.

We let go of the apple, it flows with gravity to the ground and no work is done during the fall since the apple is motionless (force free) in free fall, relative to the movement of gravity. When it hits the ground, then there is resistance that dissipates gravitational potential that came free from the environment.

What form of dissipation?

If it just hit with a thud, all in the form of impact, heat, etc... all of that is WORK in measurable joules or fractional joules of energy. That work added to our initial 1 joule of work in the lift equals more than 1 joule of work.

It is a fairy tale to say we are only getting out what we put back in. 1 joule was required to lift it because simply, 1 joule of work had to be dissipated to get the ball to 20cm. That dissipated work didn't get stored because if it got stored and with losses say 0.8 joules was stored at 20cm, that means that the math is wrong because it only took 0.2 joules to lift the ball and not 1.0 joules to a height of 20cm.

A - If the classical teaching says 1 joule of work is required to lift it 20cm and there is anything "stored" at 20cm, then the classical argument loses big time. 1 joule minus the "stored" potential is what it took to lift the apple and it didn't take 1 joule. Classical argument loses here.

B - If the classical teaching says 1 joule of work is required to lift it 20cm, that literally means that 1 joule of work was dissipated to lift the ball against gravity....ANY work after that is work that is ADDED to the 1 joule of already dissipated work meaning that that extra work plus 1 is over 1.0 COP - again the Classical argument loses big time here too.

Why? Because no matter what, when working with an open system, closed system thermodynamics don't even apply, period, end of story. Go see my 1 joule of energy thread and look at the references I posted in the last 10 messages or so. If you're honest with yourself, you will see the truth.

Any open system out of equilibrium will constantly get free input from the environment to keep doing work and little by little it winds down still adding to entropy of the universe, but MORE and MORE and MORE work is added to the system AS LONG AS A GRADIENT CAN BE MAINTAINED. Under 100% efficient since account for our input and environmental free input, there are losses. but it is OVER 1.0 COP because more work is done than what we put in.

Lift a bouncy ball and let go, that is over 1.0 COP. Not interested in mythical abstract imaginary concepts of storing potential in the ball at 20cm. Look at the mechanical truth to the event, there is no more gravitational potential available to it compared to when it was on the ground. Actually less since the higher up, the less dense gravity and the less potential available to it.

What I'm saying isn't abstract. The classical thermodynamics and any notion of storing of potential is abstract. Saying energy changes form from one to the next. That is abstract. My example is self evident for what it is and is not abstract.

Do you get it? There is no storing of gravitational potential.

The new gradient created by lifting the apple allows NEW potential to come into the system that didn't come from us lifting the ball.

There is no changing of forms of energy from one to the next. We expended and dissipated 1 joule on the lift. That energy is GONE, there is no transformation - it all dissipated. When the ball drops, that is NEW potential coming into the system. No changing forms of energy from one to the next...just creating a new gradient one after the other that allows new potential to be added over and over until the losses kill the ability to form a new gradient. Under 100% efficient. Over 1.0 COP.

Now when the ball drops, if it bounces up to 15 cm, that requires x joules of work to lift the ball to 15cm. Real measurable joules of work was dissipated to lift the ball 15cm. NONE of that ever came from our initial 1 joule that was ALREADY dissipated. If you don't believe it was dissipated, then the math is wrong because it says it takes 1 joule to lift it 20cm.

Again, it takes 1 joule to lift it 20cm.

1 joule to lift it 20cm.

1 joule is a measurement of WORK, that means 1 joule of work was done to lift it to 20cm. Not 0.9 joules, not 1.1 joules. So if we want it at 20cm, with losses against air resistance, etc... then we have to put in 1.1 joules to get 1 joule of lifting work to get it to 20cm, yes I am accounting for that concept but keeping the example simple.

But after 20cm of lift, 1 joule of work is dissipated meaning we are NEVER getting any stored potential back out of the ball and when the ball drops and bounces over and over until it stops. ALL of that work added to our 1 joule of work that was ALREADY dissipated is MORE than 1 joule of work TOTAL done.

That total work done divided by 1 joule of our completely dissipated 1 joule of lifting work equals..... OVER 1.0 COP.

All of this jives 100% with non-equilibrium thermodynamics, which is the ONLY thermodynamics that explains open systems. It is all totally permissible and doesn't violate anything. It is only a violation if the classical thermodynamics are LAWS that apply to everything. But if they want to revise them into conditions that only apply to closed systems, then nothing is violated. Just keep the laws in their own jurisdiction - it is a very, very simple concept to understand.

Rosemary Ainslie's circuit is the highest COP circuit I've ever seen that anyone can make. That coil is open to environmental input and because of the non-classical workings of the circuit, every bit of wire on the circuit is a path for potential to diverge into...not just the coil.

Understand these NET systems (non-equilibrium thermodynamic), and what energy and potential really is and this is all clear. The classical textbooks don't even properly explain either one.

Until next time...

[Aaron]

eliminate shunt negative spike ring

[witsend]

Aaron - I see it. Well done.

[witsend]

Luc, Don't know if I'm missing you or what. Please advise. Were you running the same experiment as per your earlier diagram but with two batteries? And were the batteries run per your experiment done without a load?

I'm not sure what the experiment was. But if you were running with a load then your results were extraordinary. Be glad to hear from you?

[poynt99]

Aaron,

Nice scope shots

Can't say I agree that there will be a net charge due to that negative spike though. I know you will probably note that I said there would be no net charge if we see an equal number of positive and negative undulations, but what is not visible by scoping the voltage is the current in the Gate resistor.

I'll try to see what I find in SPICE, so until then, don't get upset

Regarding your bouncing ball posts, are you saying that after lifting the ball to a height of 20cm (how much energy this took is irrelevant for the moment) there is no potential energy in the ball?

Are you saying that the conversion of kinetic energy to potential energy can never take place?

.99

[witsend]

Following quotes from MileHigh
I get it that the variables giving power is force x velocity and as applied to electric energy it's voltage x current. But am not sure how to understand your point ...'a charged inductor has current going through it, and no voltage across it's terminals'? I presume it's a misprint? In other words '... the inductor has current and voltage.'?

Here is a simple mechanical model for what this circuit does: You lay a HO scale train track across your bench and you have a boxcar that you are going to push back and forth down the track. You need to create some friction to slow the boxcar down like it is being pushed through mud. It's a thought experiment so it's thick chocolate pudding ooze. On one end of the boxcar you attach a moderately stiff spring that you can push on with your finger. Then you push on the spring at a fixed speed and watch the boxcar move through the thick ooze of the choclolate pudding. You notice the faster you push the more the spring compresses. You push for exactly three seconds at a constant velocity and then stop cold.
Got it. It's very clear.

When you stop, you notice the boxcar creep forward in the ooze as the spring decompresses against your stationary finger. Then after 5 seconds you push again. You notice the spring compressing before the boxcar starts moving, and then for a few seconds you are "plowing through the ooze" and pushing the boxcar forward at a constant speed, you stop cold and see the boxcar inch a bit further forward through the ooze as the spring decompresses again.
Again very clear.

When your finger moves forward at a constant speed, that's when the MOSFET switches on and you are providing voltage to the circuit (voltage = velocity). When your finger stops moving, the MOSFET switches off.
Still with you.

The chocolate pudding is the resistive part of the coil-resistor. When you move through the thick ooze you are exerting a lot of power, and that power is burned off as heat in the ooze, just like bending a coathanger. When you push the boxcar through the ooze, you are turning the mechanical power you are providing, force x velocity (your finger) into heat. You push your way through the ooze and the ooze heats up. i.e.; the resistor heats up.
OK

When you stop, the spring starts to decompress. Very important point here: When you stopped applying force into the spring by stopping cold, the spring now starts to decompress and keeps pushing on the boxcar with exactly the same force that you were originally pushing with. Can you see that? The force pushing the boxcar forward remains the same even though you stopped pushing. Just like an inductor will keep the same current flowing.
Still OK'ish

Now the short version: You start pushing, the spring stores some of your expended energy before the boxcar is plowing through the ooze and burning off the power you are supplying with your moving finger and heating the ooze. When you stop, the energy you put in to compress the spring is the same ernergy that you see when the spring decompresses and pushes the boxcar forward a bit through the ooze. Wait five seconds and repeat.

I can't fault the analogy and the reasoning but only if, indeed, voltage applied is the 'precursor' to the current. But that's not what Farrady and Maxwell require. The two are simultaneous - the electric and the magnetic fields being precisely equal and concurrent. In other words - while the box car is being pushed - at the same time it's pumping some sort of energy into a capacitor that it keeps inside. The precursor to current flow is potential difference. That's the finger pushing the box car. It's not voltage. It's simply the quantity of energy available should you link it to a railway line. Give it a railway system - let go some of that potential difference and you get a release of voltage and current at precisely the same time and in equal proportions. The current flow 'electric' generates a secondary supply of potential difference 'magnetic' across the box car.

[witsend]

So. That's as I see it the difference between classical and new age - if I can call it that. Classical has the concept of 'push-pull' in some sort of time sequence. New age says the 'push-pull' are simultaneous.

I think the value of the 'spike' will answer this question. If, indeed it is shown that the amount of energy in the 'spike' equates to the amount of energy applied during the 'on' period of the switching cycle - then it may be fair to conclude that energy in was also energy regenerated and returned. If, on the other hand it is found that the 'spike' bears no real relation to regenerated energy - then it would be impossible for the battery to recharge at all as that amount could never exceed the resistance from the battery supply source itself.

Poynt, I think we need your expertise here.

[witsend]

Aaron,

Regarding your bouncing ball posts, are you saying that after lifting the ball to a height of 20cm (how much energy this took is irrelevant for the moment) there is no potential energy in the ball?

Are you saying that the conversion of kinetic energy to potential energy can never take place?

.99

If I can try and answer this Poynt - I think Aaron's saying that this 'bounce' is the point where it exceeds one. Not the lifting or the letting go. If he had a stone and a ball weighing the same - the stone may not bounce. The rubber ball will.

EDIT - I see it. I haven't read his post. I just know what we've discussed. I'll check it out.