Claimed OU circuit of Rosemary Ainslie

[TinselKoala]

I'm not interested in following your bogus dog and pony show so haven't kept up on how you are misusing your equipment so there is nothing for me to explain to Rosemary.

You can't hide behind that mustache all your life. You obviously have zero confidence in anything you are doing and because you think you are anonymous, anything you post is irrelevant to this entire project.

Be a man tk and put your real name on your work here. Otherwise, it is meaningless garbage.

That's pretty funny coming from you, Err-on. You might think my work is irrelevant, and you are probably right, because it's sort of like working really hard to prove that the moon isn't made of green cheese. But as long as people like you and Rosemary are trying to sell green cheese as caviar, I'll keep working to show how ridiculous you are.

It's going to be winter soon, if it already isn't where you are. How will you heat your home?
How will you charge the batteries in your electric vehicle?
Using an Ainslie circuit?
Somehow, I don't think so.
But if it made COP>17, what's to stop you?

And you keep accusing me of being bogus. What exactly of my work is bogus? I give all details and anybody can repeat it for themselves. But you--your latest 555 diagram is even wrong.
And let's not even talk about the bogosity of zipons and antitruants, and trying to replace QED with a "theory" that cannot even make a single testable prediction--because its originator has no math beyond arithmetic.
Bogus bogus bogus. Bogus "patent", bogus Quantum circuit diagram, bogus understanding of duty cycle, bogus "random aperiodic resonant Hartley oscillations", bogus bogus bogus.
Dog and pony show? The only thing I have trouble with is figuring out which one of you is the dog and which is the pony.

[qiman]

From Rosemary:

TK - I'm hoping I can bend your mind around this problem.

The battery recharges, power through, voltage first drops - then a spike to, what was it - say 50 volts or thereby? At that same moment the value across the shunt say 0.4volts positive drops to about 1.2volts negative, (aproximate because I couldn't see the actual value) . Then how do you work out the product of the energy available at that moment. In my reckoning it is 1.2/0.25 = plus/minus 4.8 amps. So. I need to be reasonably certain that the actual energy calculated at that moment as v*i = 240 watts BACK TO THE SYSTEM. (Again not shouting. Just emphasis)

Now. If you do not do the integration simultaneously how is this advantage or gain made evident? And more alarmingly, if you simply do the product of both values and add it to the general loss - then your methodology is wrong.

If, as you say the LeCroy is doing the math - then I'm afraid I need to see it. I'm still concerned that it is not capable of doing that DC offset. To my primitive way of thinking this means that it cannot gauge zero in order to make a comparison as to where the waveform was found in relation to zero.

Please explain this. Sorry to impose.

You are right. Its's the best I can do to explain my concern. I use primative example and analogy. But it serves its purpose.

[exnihiloest]

...You have tried to describe this process in your primitive way many times apparently without realizing you are describing a standard procedure and one that is well understood--integration.That's right, you cannot understand it. The little Fluke cannot do the integration; the operation must be done on the power waveform NOT on either the voltage or the current waveform individually. That is another error you seem to be making. The voltage and current must be multiplied together first, THEN the integration is done. If you are doing it on the voltage first, you are making another error.
...

That is right. dW=U(t)*I(t)*dt and an integral of a product is not generally the product of integrals.
To integrate separately voltage and current is an error of beginners in electronics and ignorants in physics.

[TinselKoala]

The battery recharges, power through, voltage first drops - then a spike to, what was it - say 50 volts or thereby? At that same moment the value across the shunt say 0.4volts positive drops to about 1.2volts negative, (aproximate because I couldn't see the actual value) . Then how do you work out the product of the energy available at that moment.

The term "product of the energy available at that moment" is nonsense. A "product" is the answer you get when you multiply two numbers. What happens here is that the INSTANTANEOUS CURRENT VALUE (say, at a time t=0.0000000001 sec) is multiplied by the INSTANTANEOUS VOLTAGE VALUE at that SAME TIME, t=0.0000000001 sec, to give a single value of the INSTANTANEOUS POWER. Power, not energy.  And it's not "worked out", it's calculated by the powerful and expensive computer in the LeCroy. And it's an entire waveform, not just a single value. Every one of the 500,000,000 samples per second is used in the calculation.

In my reckoning it is 1.2/0.25 = plus/minus 4.8 amps. So. I need to be reasonably certain that the actual energy calculated at that moment as v*i = 240 watts BACK TO THE SYSTEM. (Again not shouting. Just emphasis)

If the voltage drop across the shunt at a given instant is 1.2 Volts, then the current  in the shunt is given by 1.2/0.25, which is 4.8 Amperes, INSTANTANEOUS CURRENT. The instantaneous power dissipated in the shunt is then  ( 4.8 x 4.8 )( 0.025 ) or about 6 Watts. This represents POWER DISSIPATED IN THE SHUNT, not "going back" anywhere, and to find the energy this instantaneous power waveform must be integrated over a suitable time period. The number you quote is an instantaneous power value for a peak of the power waveform In the Circuit, whose resistance is somewhat more, and is offset as you know by a similar peak in the other direction. This is what DrStiffler and others mean when they say "The peaks integrate to zero". That is, they nearly perfectly cancel each other out, WHEN THE WAVEFORM IS INTEGRATED to produce the energy value at a given time.

Now. If you do not do the integration simultaneously how is this advantage or gain made evident? And more alarmingly, if you simply do the product of both values and add it to the general loss - then your methodology is wrong.

Now, you finally (almost) have added "integration" to your vocabulary. Unfortunately your understanding is still incomplete--as it will remain until you study the calculus. Integration is a process that is "outside" of time in a way, since it adds up all the time slices of a waveform to figure the area underneath it. Energy is the PRODUCT (two numbers multiplied, remember?) of POWER and TIME. Power is Watts, Energy is Joules, Joules are Watt-seconds. So if you have a GRAPH of POWER vs. time, the AREA under the curve on that graph is Power times Time, which is Joules, or energy. Integration operates on an entire waveform at one time and gives the area under the waveform __FULLY ACCOUNTING FOR THE POSITIVE AND NEGATIVE SWINGS_ (that is a SHOUT).
My methodology is not wrong. Your attitude and lack of understanding and your arrogance are what is wrong.

If, as you say the LeCroy is doing the math - then I'm afraid I need to see it. I'm still concerned that it is not capable of doing that DC offset. To my primitive way of thinking this means that it cannot gauge zero in order to make a comparison as to where the waveform was found in relation to zero.

IF, as I say? Who's doing the math if it isn't the LeCroy? Certainly not you.
Can you see the math inside your computer?
As I have TRIED to explain to you over and over, the DC offset issue affects ONE of the scope's TWO input channels. It does NOT affect any of the scope's other functions and has been inactive, that is, I have NOT used it at any time in these demos or calculations. It has nothing to do with the TRACES or the MATH displayed, and if you had any real inkling about scopes you would know this. You are tilting at a non-existent windmill. Your primitive way of thinking is in error, as even Aaron should be able to explain to you.

Please explain this. Sorry to impose.

You are right. Its's the best I can do to explain my concern. I use primative example and analogy. But it serves its purpose.

It IS an imposition, because I am trying to explain to you the things that are normally covered in a high-school precalculus class lasting three months, where students actually DO HOMEWORK and SOLVE PROBLEMS like the one we are trying to discuss here. And this kind of math is considered PREREQUISITE for any even elementary study in electronics. So I am doing remedial math and EE with someone who thinks she knows it all already yet is making basic errors with almost every step, and compounding them with willful misunderstanding--hence I know my effort is futile.
And you are NOT sorry at all.
But at least some others reading here will know what's up.

[TinselKoala]

You can take that 240 Watts you calculated, and mutiply it by the time width of the spike that produced it.

The frequency of the spikes in my rig is about 1.6 Mhz, so a single cycle occupies 1/1,600,000 second.

Approximating the spike by a long skinny rectangle (which OVERESTIMATES it) we get 150 MicroJoules of energy in the spike. This is offset by another, say, 140 microJoules coming back in the negative part of the cycle. The difference, 10 millionths of a Joule per cycle, is dissipated as heat.

This can be clearly seen in the LeCroy's display, as the Energy integral has "humps" in it where the spikes and ringdown are occurring--the energy (flowing one way) stops increasing and decreases a bit (representing the reverse flow) before resuming its normal course.

The integrated Power trace--that is, the energy-- on the LeCroy clearly shows that small reverse energy flow--and also clearly shows that it does NOT accumulate in the battery, it merely puts a tiny amount of the battery's energy back into the battery--or it can be made to accumulate into a second batt or cap external to the circuit. This energy comes from the running battery and the total is NOT increasing.

It's easy to calculate the energy that can be returned to the battery by this method. You can even neglect the dissipation due to heating and just say that the entire spike decay is due to energy passing back into the battery.

So figure 150 microJoules is actually going back to the battery on each pulse of the primary freq--which is 2.4 kHz. So each second there is ( 0.00015 ) x (2400) or 0.36 Joule going back.
Really there is much less, but OK. So that's 0.36 Joules per second going back. 0.36 Joules is 0.36 Watt-seconds, and 0.36 Watt-seconds per second is 0.36 Watts, AVERAGE POWER that may be returning to the battery. And that's an overestimate from several overestimated approximations.

The average power being dissipated in the load, meanwhile, is something like 2.5 or 3 Watts, or nearly eight or ten times as much. So about 10 or 15 percent, under the very most favorable assumptions, is available to get back to the battery. As the LeCroy shows, the actual number is much smaller, more like 2 percent.

Even using the numbers you come up with, Rosemary, you can see that the actual average power available to return to the battery is quite small. Most of this 0.36 Watts is actually dissipated as heat.

The real numbers can be read directly (using a simple conversion factor) from the display of the LeCroy.