COP 17 Heater | Rosemary Ainslie

[poynt99]

Shunt Resistor Location

One thing I would like to suggest Rosemary, is that unless you specifically require that the 0.25 Ohm shunt resistor be placed in series with the MOSFET Source for some good reason, I strongly suggest that it be relocated to be in series with the battery.

Having that resistor, even as small as it is, affects the operation of the MOSFET switch, especially if that resistor is somewhat inductive, although it should not be if it is a carbon resistor, which I believe you are using. Nonetheless, it is not good practice to place the resistor there if all you want to do is monitor current consumption from the battery.

Please specify why you require it there in the MOSFET Source instead of in series with the battery?

.99

[witsend]

Ok. I guess we're not going to break new ground here. Just please bear in mind that neither of you have addressed the question. So let me put it back on the table and, when you're ready, maybe you could consider it. This, after all - is the moment that has puzzled those experts who have dealt with this problem and had the intellectual honesty to ask our academics to please address it. That was the basis of their accreditation. In other words they identified the problem. They wanted some answers.

What happens to the potential difference on that inductive resistor when the battery can no longer deliver energy?

And Poynt, I really can't argue against simulated programmes. But I can point to the experimental evidence. Take another look at the paper.

And both of you, MileHigh and Poynt, many thanks for your input here. It's an interesting discussion. But it needs a more critical overview than conventional physics allows.

EDIT - I was rather hoping for acknowledgement here from you Poynt. By which I mean, I actually thought you were prepared to at least look at the bare bones of the argument.

[poynt99]

Shunt in Series with the Battery

Here is a nice scope shot of the coil current (green) actually reversing (unless MH has a different explanation), but no current going back into the battery (red trace).

This is with the current shunt in series with the battery.

.99

Edit: I've added a shot of the flyback diode current (green) vs the MOSFET Gate drive voltage (red).

[witsend]

One thing I would like to suggest Rosemary, is that unless you specifically require that the 0.25 Ohm shunt resistor be placed in series with the MOSFET Source for some good reason, I strongly suggest that it be relocated to be in series with the battery. Poynt99

It makes not a blind bit of difference where we put the shunt. But the shunt - plus the probe connections plus the solder for the shunt - all interfere with the resonance. Not critically - but some. Therefore it is better at the negative rail. And, as I've told you, the idea of the test is to encourage resonance. Why insist we position it where it might interefere?

But also - as mentioned - our accreditors had the shunt in many many more positions than the 1 shown in our circuit. Just depends where one actually wants to see the current flow.

[witsend]

Here is a nice scope shot of the coil current (green) actually reversing (unless MH has a different explanation), but no current going back into the battery (red trace).

This is with the current shunt in series with the battery.

.99

With respect. I have no intention of taking any of TK's probe shots seriously. I just am not prepared to go into my justifications here. If you prefer his experimental evidence - then that's your choice.
EDIT - Quite apart from which - put the shunt on the positive and the negative simultaneously - it will show the same spike at the same level. So what does this prove? That it has been shuffled around the circuit and never went near the battery? How odd.

[poynt99]

Good grief Rosemary,

These are MY scope shots from my simulation. I did say "sim" in my post.

If you are trying to make the circuit unstable and more "noisy", then putting the shunt in the Source is the way to go.

You have to give us a bit of patience here as we designers always try to use good practices to minimize noise etc., not enhance it.

If it makes no difference then that is my recommendation for any future testing, including replications here.

.99

[witsend]

Good grief Rosemary,

These are MY scope shots from my simulation. I did say "sim" in my post.

.99

Apologies. I'm way too prickly. Ok. If the simultor shows the spike ask the simultor if it could have come from the coil? EDIT or better still - ask the simulator what comes from the residual PD at the coil? Poynt - that's your real expertise. Do you make allowance for this moment?

EDIT another poynt. The parasitic Hartley Effect - is easily snuffed. It takes a little bit of pressure. That's why it's better to keep that rail clean. These effects, when encouraged seem to add to the benefit in recharging batteries to add to overall efficiency.

[witsend]

Guys I'm exhausted. I'll take this up again after a bit of shut eye. It's been a really good chat. Thanks. I think we're finally getting to the heart of the matter.

Thanks
Rosemary

[Aaron]

Resistive Inductors

From John Bedini's page:

I have set up this page you may review what has been going on in the past three years

Let's look at it this way, If we go back to where Tesla said that when the switch was thrown in the DC power station the operator was killed by some unknown! Tesla studies this , He reproduces it in his lab where he says that a great stinging pain was felt by him ( the gaseous fluid, not controllable passes through everything). He finds out that just before the current carriers catch up and enters the highly resistive inductor (by sharp repetition rate) their is a radiant pulse that causes the wire to emit a glow that takes place, then the current catches up and it goes away (normal positive EM power). Then the current is cut off and a back EMF develops. These are two different things! In my tests I have found that the radiant only appears to traverse the coil when voltage appears, in other words this radiant pulse is there when the voltage potential leads the switching, then the current makes it's way through the resistive coil. But now the switch is off and the back EMF shows up only reverse at 1/3 the radiant pulse, so in a sense, you could say this all takes place unidirectional in few microseconds of time.

Gabriel Kron

We quote: "...the missing concept of "open-paths" (the dual of "closed-paths") was discovered, in which currents could be made to flow in branches that lie between any set of two nodes. (Previously — following Maxwell — engineers tied all of their open-paths to a single datum point, the 'ground'). That discovery of open-paths established a second rectangular transformation matrix... which created 'lamellar' currents..." "A network with the simultaneous presence of both closed and open paths was the answer to the author's years-long search."

[Aaron]

scope shot @ 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

Thanks but at least is is undeniable that whether or not there is a ring in the negative spike, it can be dealt with perfectly.

Now that there is a legitimate spike, undeniably, the true RMS power reading from a scope will tell the story.

---------------------------------------------------------------

off topic - but relative to the concept of charging coil and discharging coil in the Anislie circuit from my perspective.

When the ball is at 20cm, tell me what is being "stored"?

What change in the intrinsic nature of the ball has happened?

When a ball is sitting at 20cm - SITTING at 20cm, compared to the ground it is in equilibrium with its environment. 100% of the gravitational potential is being dissipated by the resistance to free fall.

When the ball is pushed off the shelf or let go...only then is the gravitational potential not 100% dissipated anymore. And is is ONLY then, during the actual fall that the free potential from gravity is available to the ball as potential.

Sitting still and ANY height, 100% of gravitational potential is dissipated by the resistance keeping it from free fall.

Holding it at 20cm before dropping it, same story. All the gravitational potential is dissipated by the resistance against free fall by your two fingers holding the ball.

Only when the ball has no resistance to gravity is there ever gravitational potential available to it that isn't being dissipated. On the way down, some is dissipated against air resistance, etc... and that potential that is ONLY available to it while it is moving down with the flow of gravity will move from this high potential to a low potential when it encounters any resistance. Impact on ground, etc... The impact is heat...the gravitational potential meeting resistance dissipates and that is work.

At the "subatomic" level, at a higher height, there is LESS gravitational potential actually available to it. The higher an object, the LESS gravitational potential available to it anyway.

An object at the peak of the highest mountain weighs less than it does at ground level. That is because there is less gravitational potential to push down on the mass of the object.

There is no such thing as storing gravitational potential.

You mention kinetic energy to potential energy.

Are you talking kinetic energy of the ball moving up to the height? Because that is kinetic ENERGY since energy is work.

Or are you talking about kinetic potential (not work) on the way down. That is kinetic potential and relative to the movement of gravity, the ball is MOTIONLESS. It isn't moving in free fall relative to gravity. That kinetic potential is gravitational potential, which turns to work when encountering resistance that dissipates it.

So on the way up, if you want to tell me the math says 1 joule of work is expended to lift the ball to 20cm. Then the argument must be consistent that if it takes 1 joule of work, that is 1 joule of dissipation. There is nothing to be stored. At 20cm, when the ball is sitting, where is this abstract "potential" coming from? It isn't stored in the apple because that concept is a figment of the imagination. It is in equilibrium with no stored potential. Let is go, then gravitational potential that did NOT come from us is pushing the ball down.

That is NOT our kinetic energy input turning into potential energy. The actual substance of the potential came from outside of our input.

[Aaron]

voltage leads current

The precursor to current flow is potential difference.

Exactly! Voltage ALWAYS leads current. Current is slow to catch up relative to voltage speed that is practically instantaneous. Supposed to be limited to about light speed anyway.

[Aaron]

rock

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.

Right. But even the rock will generate heat on impact and that is measurable work. Still over 1.0 - heat work added to our lift work.

Of course that heat work isn't usable work to do anything constructive but it is work.

If there is a spring loaded mechanism the rock fell on, it would compress the spring and be bounced. So instead of doing all heat work, there is less heat work and some lift work. Over 1.0 again.

[Aaron]

no stored potential at a height

When the ball is at its high point it does not have stored energy, in fact it has no energy at all, the energy comes from gravity when it starts to fall and when it hits the floor this is when it stores its energy so as to counter gravity to rise up again.

A battery stores energy which was put into it in the first place, and to put this energy in a change took place in the battery, the ball did not change when it was at its high point.

Beautiful!

[Michael John Nunnerley]

Well Agree To Disagree

I see.

Well, good luck with that.



.99

Well thanks for the arguement, I like a good arguement especially when I am on the right side I had an arguement with my government some time ago, but they paid, at the time, but now I am doing the same thing but in a different way as to the agreement, and with a lot of security, apart from I am now a lot older and my days or years are a lot less and so I do not worry like I did before.

My name is as you see it in this forum, I do not hide, everything that I say is the truth as I find it, I am not always right, I admit to that, but anybody that knows me personally knows I have the background to suport the things that I know and say.

I have been working for decades on renewable energy and I have a lot of enemies in this field, but also a lot of friends. Now are you friend or foe?

Mike

[Aaron]

bouncing ball

Each time the ball bounces you loose a little puff of energy in the deforming ball, and hence the voltage across the ball drops. So the bouncing ball's voltage takes multiple discrete steps down in time, down to zero.

Yes, of course the ball looses energy after each bounce. I have never suggested the ball being let go of will be able to bounce higher that from where it was released during the next single bounce. But....

I drop it at 36 inches for example, before it stops, it will have bounced a combined total of about 96 inches in upwards lift adding all the upwards bounces. I know because I did the experiment with the ball next to a tape measure and I measured the combined inches of lift from the time I let go. That is 3 times more lift than the math says I should get from the input of lifting it 36 inches.

Non-equilibrium systems don't mean over 100% efficient, which is what your argument is based on. But it does mean over 1.0 COP, which your argument doesn't address.

These open system concepts is simply allowing dissipation to happen slower over a longer period of time allowing simply more work to be done before all the potential is dissipated.

It isn't magical and it isn't perpetual motion. It is free energy from the environment giving us what we need to do what we need to do because this is a universe of ABUNDANCE and not a universe of poverty.

This all plays into the coil charging and discharging.