Checking out Joel Lagace - Hang on I'M SERIOUS!!

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Magluvin

Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #200 on: 2025-03-06, 00:13:57 »

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Quote from: verpies on 2025-03-05, 23:26:34

This oscillogram represents a slowed-down view of the energy transfer process and the voltages levels you are describing are not occurring in this circuit.
Are you blind ?
In this circuit, the caps are never both at 7.07V (initially, C1 decreases from 10V to 0V and only after that C2 increases from 0V to 9.99V).
Also, the inductor's current (I_L1) is at maximum when both of the caps are empty (at 0V, not at 7.07V).
This oscillogram illustrates that unambigously.
Don't change the subject. I am not referring to a circuit connected like the one you are describing.
In all my previous messages, I linked this circuit so there would be no doubts to what I am referring to.
Exactly.

firstly, dont change the subject?

your circuit is not representative of what we are discussing. so you can shut it.

we are talking about cap to cap. trying to use the cap to cap paradox and seeing what we might be able to do with that. this is the subject of the thread and joels claim as such. no power supply connected except to initially charge cap 1. so, just 1 charged cap and 1 completely discharged cap with some simple circuitry, to try and charge a 3rd cap with the energy transfer from cap1 to cap2.

so if you are not talking about cap to cap as we are, then should i even respond to the rest of your post?

mags

verpies

Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #201 on: 2025-03-06, 00:33:23 »

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Quote from: Magluvin on 2025-03-06, 00:13:57

firstly, dont change the subject? your circuit is not representative of what we are discussing. so you can shut it.

It is related. Did you see what I wrote below ?

Quote from: verpies on 2025-03-05, 17:47:50

Yes, cap-->cap energy transfers are inherently inefficient.
This is because caps are voltage sources, yet for the most efficient charging process, they require a current source.
This is also the reason why coil-->cap energy transfers are so efficient (coils ARE current sources).

It is possible to transfer 99% of the cap's energy to another cap by using a low-loss coil as an intermediary, i.e.: cap-->coil-->cap transfer (C1 --> L1 --> C2 in the circuit below).

...and you engaged with that specific circuit by replying:

Quote from: Magluvin on 2025-03-05, 18:57:28

the inductor used to transfer most or all of cap1 to cap2 would need to be absolutely huge if using super caps, if not virtually impossible.

Thus there could have been no doubt what we were discussing.

« Last Edit: 2025-03-06, 01:38:36 by verpies »

Magluvin	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #202 on: 2025-03-06, 01:19:43 »
Group: Mad Scientist Hero Member Posts: 676	if we do as joel has shown, he is intentionally letting 50% of the original energy from cap1 by way of the cap to cap paradox, in hopes of. recovering it, or better to cap3. anyway, if i charge cap1 to 100v and direct discharge that into cap2, both caps same value and cap2 stars at 0v, how much voltage will be in each cap when all is said and done? why, in the process described did we lose 50% of that energy? mags

verpies	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #203 on: 2025-03-06, 01:29:30 »
Group: Professor Hero Member Posts: 3670	Quote from: Magluvin on 2025-03-06, 01:19:43 why, in the process described did we lose 50% of that energy? To i²R losses, arcing and EM radiation. The rate of change of current approaches infinity when two unequally charged caps are connected because caps are voltage sources, and for the most efficient charging process, they require to be charged from a current source.

Magluvin	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #204 on: 2025-03-06, 01:44:36 »
Group: Mad Scientist Hero Member Posts: 676	ok. im looking at your circuit. it is using just an inductor instead of a transformer. i dig it. im on a similar path on that. you have pretty much applied a boost converter to do cap to cap. i see your scope shot. yes you charge the inductor with cap1 and then discharge that into cap 2. thats why you have both caps at 0v before the inductor charges cap2. joel is not doing it that way at all. he is just pulsing the primary with cap to cap currents in pulses now that there is consensus that what joel has presented camnot work as presented, i am still in the process of replacing the transfomer with my orbo. i intend to show transfer of cap1 to cap2 all in one shot and charge a cap3, as the input of the orbo, a toroid winding, is not affected by the output winding. rewinding the toroid winding as we speak for higher inductance so i can do larger caps that dont self discharge quickly like the 2uf i found good results with. ive explained all this earlier. circuit is already set up. as soon as i fin the winding it will be ready to go. basically i will be doing what you are showing but all in 1 shot. cap1 to cap2 and output to cap3. it will be a proof of concept mags

Magluvin	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #205 on: 2025-03-06, 01:49:48 »
Group: Mad Scientist Hero Member Posts: 676	Quote from: verpies on 2025-03-06, 01:29:30 To i²R losses, arcing and EM radiation. The rate of change of current approaches infinity when two unequally charged caps are connected because caps are voltage sources, and for the most efficient charging process, they require to be charged from a current source. i dont believe weblost the 50% due to resistance nor radiation. cant explain it again right now, but i have explained it many different ways here already. check it out. tell me what you think excuse some typing errors. from my phone mags

Itsu

Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #206 on: 2025-03-06, 09:30:30 »

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Quote from: verpies on 2025-03-05, 17:47:50

Yes, cap-->cap energy transfers are inherently inefficient.
This is because caps are voltage sources, yet for the most efficient charging process, they require a current source.
This is also the reason why coil-->cap energy transfers are so efficient (coils ARE current sources).

It is possible to transfer 99% of the cap's energy to another cap by using a low-loss coil as an intermediary, i.e.: cap-->coil-->cap transfer (C1 --> L1 --> C2 in the circuit below).
In such scheme, the switch and diode/synch rectifier play a role of gatekeepers that do not allow the energy to return to the first cap and coil.

To see it working click here.
The op-amp's/comparator's purpose is to detect when C1 is empty.
The R1 is there only to make the simulation well-behaved (in reality it can be omitted).
The MOSFET'ized version is here

Interesting topic this "cap to cap transfer via and inductor".

I was looking into it too after your response and found this on the net: https://www.av8n.com/physics/capacitor-transfer.htm

I was trying to do an LTspice simulation, but got a bad case of the flu i am now dealing with.

Also ChatGPT "knows" about it, but its diagrams sucks.

So we basically dump the C1 charge into L1 (the diode prevents C2 from being charged initially), then at the negative slope, C2 gets charged up from the energy stored in L1.

Itsu

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

Checking out Joel Lagace - Hang on I'M SERIOUS!!

CLC Xfer Graph.png (15.7 kB, 730x390 - viewed 148 times.)

verpies	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #207 on: 2025-03-06, 09:38:31 »
Group: Professor Hero Member Posts: 3670	Quote from: Itsu on 2025-03-06, 09:30:30 got a bad case of the flu i am now dealing with. I had that flu, too. It is bad and lasts for weeks. Treat it seriously. There are also 3 other virii going around. See: https://youtu.be/oHU4LBdTl0o?t=234 Quote from: Itsu on 2025-03-06, 09:30:30 So we basically dump the C1 charge into L1 (the diode prevents C2 from being charged initially), then at the negative slope, C2 gets charged up from the energy stored in L1. Yes, but it is possible to do that transfer without that diode and without the loss due to its forward voltage drop. Just with one MOSFET ...but the circuit for that is less educational.

verpies	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #208 on: 2025-03-06, 09:44:23 »
Group: Professor Hero Member Posts: 3670	Quote from: Magluvin on 2025-03-06, 01:49:48 i dont believe we lost the 50% due to resistance nor radiation. But you do acknowledge that EM and i²R losses happen during a CRC energy transfer (not my circuit!) qualitatively, don't you ? If you do, then all that remains is to establish the magnitude of that loss quantitatively. For example: for two equal caps (C₁ and C₂), initially one charged to 10V and the second to 0V, the current flowing between them in the first instance when they are connected by a 100mΩ resistance (R) is 100A and that represents 1kW of power loss in that instance (because P=i²R). To obtain the total energy loss this power loss needs to be integrated over time as the current falls^. The result obtained this way beats any belief you may have about it. This does not consider the energy loss in the EM radiation resistance. The current in a series CRC circuit falls in time like this: i(t)=[(V₁-V₂)/R]e^[-t/(RC₁C₂/(C₁+C₂))] Integrating the instantaneous power loss Ri(t)² over t from zero to infinity, yields the total energy loss as E=[C₁C₂(V₁-V₂)²]/[2(C₁+C₂)]. Note that E does not depend on R. If C₁=C₂=C and V₂=0 then the above reduces to: E=C²V₁²/4C = CV₁²/4 = ½CV₁²/2

F6FLT

Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #209 on: 2025-03-06, 11:36:02 »

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Quote from: Magluvin on 2025-03-06, 01:49:48

i dont believe weblost the 50% due to resistance nor radiation.
...

Why don't you believe what all electronics engineers know and use every day?
You waste half the energy charging a capacitor through a resistor, from a voltage source like another capacitor or a battery.

Since the voltage of the capacitor and the voltage source are different as long as the capacitor is not charged, the current would be infinite. In practice, since there will always be a resistance in the connection and in the capacitor itself, the current peak will be dissipated in the resistance and will represent an energy equal to that of the capacitor charge.

The only way to charge a capacitor without loss is with a current generator, not a voltage generator. A current generator allows the voltage across the capacitor to rise step by step.
Alternatively, a voltage generator can be used, but must be connected in series with an inductor of sufficient value, which transforms it into a current generator by the virtue that E=-L.di/dt (so E will also rise smoothly).

Radiation loss will only be significant if you use coils of not negligible size in relation to the wavelength of the frequencies in the charge current spectrum.

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

"Open your mind, but not like a trash bin"

Magluvin	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #210 on: 2025-03-06, 16:26:31 »
Group: Mad Scientist Hero Member Posts: 676	Quote from: F6FLT on 2025-03-06, 11:36:02 Why don't you believe what all electronics engineers know and use every day? You waste half the energy charging a capacitor through a resistor, from a voltage source like another capacitor or a battery. Since the voltage of the capacitor and the voltage source are different as long as the capacitor is not charged, the current would be infinite. In practice, since there will always be a resistance in the connection and in the capacitor itself, the current peak will be dissipated in the resistance and will represent an energy equal to that of the capacitor charge. The only way to charge a capacitor without loss is with a current generator, not a voltage generator. A current generator allows the voltage across the capacitor to rise step by step. Alternatively, a voltage generator can be used, but must be connected in series with an inductor of sufficient value, which transforms it into a current generator by the virtue that E=-L.di/dt (so E will also rise smoothly). Radiation loss will only be significant if you use coils of not negligible size in relation to the wavelength of the frequencies in the charge current spectrum. short on time atm. but if you look back on this thread, ive gone over reasons why i 'claim' what im what im talking about. ill be back on later. at work real quick, on losing 50% by charging a cap from a power supply, or other stable source. thinking on that as i have my mind set, that loss may not be a loss. ill think today and let u know what i think about that. im hoping that even one of you get what im claiming. talk later thanks mags

web000x

Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #211 on: 2025-03-06, 19:49:40 »

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Posts: 340

Quote from: Magluvin on 2025-03-06, 16:26:31

im hoping that even one of you get what im claiming. talk later

thanks

mags

I got lost on your theory when you started talking about a cap transfer/discharge where it was an ideal cap with no inductance in the conductor medium. This can never be. Just as there can never be a coil that is without inter-turn capacitance.

Dave

Itsu	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #212 on: 2025-03-06, 20:36:37 »
Group: Elite Experimentalist Hero Member Posts: 4236	As mentioned earlier, as was trying to complete an LTspice simulation with a similar circuit as verpies used. His circuit and result can be seen here: My LTspice simulation result is here: As can be seen, the results are very similar where C1 starts with 10V and ends with 0V while C2 starts with 0V and ends with -10V. But -10V on an electrolytic cap would be not a very stable situation, so i am not sure how to overcome this (if still using electrolytic caps). Another observation is that after some 31 seconds or so the voltage on C2 is back to 0V, see this long time simulation: Not sure if this in real life too or a glitch (or the fact of the reversed polarity on the electrolytic cap C2). LTspice file attached Itsu ------------------------ CapacitorParadox plus inductor.asc (1.66 kB - downloaded 15 times.) Checking out Joel Lagace - Hang on I'M SERIOUS!! verpies simulation small.png (38.16 kB, 823x440 - viewed 233 times.) Checking out Joel Lagace - Hang on I'M SERIOUS!! cap paradox plus inductor small.png (65.26 kB, 950x432 - viewed 234 times.) Checking out Joel Lagace - Hang on I'M SERIOUS!! cap paradox plus inductor after 31 seconds small.png (52.12 kB, 952x436 - viewed 221 times.)

verpies	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #213 on: 2025-03-06, 21:34:14 »
Group: Professor Hero Member Posts: 3670	Quote from: Itsu on 2025-03-06, 20:36:37 But -10V on an electrolytic cap would be not a very stable situation, so i am not sure how to overcome this (if still using electrolytic caps). Just flip that cap around. It always experiences one polarity of voltage, anyway. Quote from: Itsu on 2025-03-06, 20:36:37 Another observation is that after some 31 seconds or so the voltage on C2 is back to 0V, see this long time simulation: Not sure if this in real life too or a glitch (or the fact of the reversed polarity on the electrolytic cap C2). Maybe the diode is leaking or LTSpice is really good and can model the leaky behavior of electrolytic caps biased in reverse.

Aking.21

Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #214 on: 2025-03-07, 00:31:28 »

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Posts: 489

Lagace latest.
https://www.youtube.com/watch?v=VgFkKkYYmjs

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Electrostatic induction: Put a 1KV charge on 1 plate of a capacitor. What does the environment do to the 2nd plate?

Magluvin	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #215 on: 2025-03-07, 01:22:53 »
Group: Mad Scientist Hero Member Posts: 676	Quote from: web000x on 2025-03-06, 19:49:40 I got lost on your theory when you started talking about a cap transfer/discharge where it was an ideal cap with no inductance in the conductor medium. This can never be. Just as there can never be a coil that is without inter-turn capacitance. Dave iinteresting that you say that about how i lost you when getting into the ideal versions of cap to cap. its interesting because mile high back in the day specifically brought the ideal situation to the table while i was having doubts in the back of my head on cap to cap and the consensus of why the 50% losses. so after he did communicate what he thought the outcome would be in ideal land, it hit me.... so i ran a sim with no resistance applied to any components. 50% loss.... what really led me to believe as i do after that, mile high said that in ideal land that doing a direct cap to cap, ideal, no resistance, that we would end up with 7.07v in each cap from a start of 10v. thats when the thoughts in the back of my head were starting to pan out as to why i felt there was a problem here that we are not paying attn too.. and i found my answers in coulombs law. firstly, if you think that instilling a charge in a cap is not due to electrons being taken from the pos plate and also electrons being pumped into the neg plate in order to quantify that charge, then we probably wont get anywhere with this. but if you do, then follow me here... basically if you inderstand what coulomb tells us then if we were to have a cap, no charge, and if we could take 1 electron from the pos plate and pumped 1 into the neg plate, that we could calc the exact voltage that the cap is charged to as long as we know the exact value of the capacitance. more electrons taken from the pos plate and sent to the neg plate, we can again calc the exact voltage potential in that cap with these 2 numbers, the electron count and the cap value. so also. with a known voltage on the cap and its value of capacitance, we could calc the number of electrons displaced from the pos plate to the neg plate. now. earlier i said that there would be an equalibrium of 7.07v in each cap starting with 10v more to see who would respond to that. it is an impossiblility to have 7.07v in cap1 and 2 by way of just cap to cap, no inductor. just as some here have pointed out. and we can agree on that. what milehigh claimed with the ideal example and said we would end up with 7.07v in each cap, that was the clincher to be able to solidly say that resistance is not the cause of the loss. the only way, cap to cap, for the results of 7.07v to be left in each cap would be to add more electrons to the system than we started with.. like a 10gal bucket of water to an empty 10gal bucket, we can not end with 7.07gal in each bucket. 10lb air pressure in a tank and another empty tank. we can only end up with 5lb in each when said and done. in order to end with 7.07lb in each tank, more air would need to be added to the closed system. same for the water.... if you use verpies circuit, and started with 10v, and he turned off the transistor when cap1 got down to 7.07v, that inductor will charge cap2 to 7.07v. again, minus diode drops. total energy left idealy in cap1 and cap2 would be equal to the energy we started with in cap1 at 10v. like mile high tried to be slick with the ideal example that he claimed we would end up with 7.07v in each cap if it were ideal, of which would show no loss. well the only way that could happen is if the electron count for each cap were altered from an external source durring the process. cant happen. my point is that if we do cap to cap, and were to calc how many electrons were taken from cap1's pos plate and put in the neg plate in order to read 10v, once we do the cap to cap, and end up with 5v in each cap, we can calculate the number of excess electrons on each neg plate and how many are depleted from the pos plates, and those calculations will show numbers that we did not add nor lose any electrons to the closed system of the circuit durring that cap to cap event. we only reduced the initial energy we began with because we didnt do anything with that transfer from cap1 to cap 2. resistance of any value, even absolute 0ohms, the outcome will be exactly the same. the only difference between lower or higher resistance would be the time it takes to do the full conversion of cap1 at 10v, to cap1 and cap 2 each having 5v. now with the inductor, things change up. milehigh said that the inductor negates the resistance in order to end up with 10v in cap2 from a start of 10v in cap1.. i was like, WHAT??? Resistance is no longer a problem what so ever now? na. makes no sence. meanwhile, with no inductor, any resistance gives the exact same results, 50% loss every time?? any other circuit, add a series resistance and change that resistance and different results will occur for each change. but with cap to cap the ONLY thing that changes is the time. thats it. why nobody elese questions this is beyond me. and once i convince some that the resistance is not the cause of the loss, they always then say that the cause of the loss is due to radiations. if that were true, then radiation losses of 50% would simply happen in everything we do. cap to cap with inductor and diode has no radiation losses??? give me a break. something is wrong when it comesnto the science on this subject. also, i dont think we lose 50% when charging a cap from a source like a power supply, etc. if you do, can you explain that to me? and i will respond. there is a BIG difference between a constant source and a charged cap as a source, on that is definitely not constant at any time durring the transfer, just a continous depletion of potential from start to finish. mags ------------------------ Checking out Joel Lagace - Hang on I'M SERIOUS!! 20250306_184942.jpg (3546.9 kB, 4080x3060 - viewed 126 times.)

verpies	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #216 on: 2025-03-07, 03:33:01 »
Group: Professor Hero Member Posts: 3670	Quote from: Magluvin on 2025-03-07, 01:22:53 what really led me to believe as i do after that, mile high said that in ideal land that doing a direct cap to cap, ideal, no resistance, that we would end up with 7.07v in each cap from a start of 10v. That was only MileHigh's hypothetical example, based on the premise: "WHAT IF there were no losses during C2C energy transfers" and the mathematical identity: 0.707...² = ½ which implies that ~70.7% of the voltage represents 50% of the energy, because the energy stored in a capacitor is proportional to its voltage squared (V²). In reality this does not happen because there are always at least 50% energy losses during C2C transfers and when the energy in two equal capacitors equalizes then each capacitor stores only ¼ of the initial energy,. This means that only 50% of the initial voltage is left in each capacitor ( not ~70.7% ). This happens because ½ * ½ = ¼ and the energy stored in a capacitor is proportional to its voltage squared (V²) thus 50% of the voltage represents 25% of the initial energy. Reiterating the relationship between voltage and energy stored in a capacitor: ~70.7% of voltage = 50% of energy. 50% of voltage = 25% of energy There is no charge conservation violation with the true numbers for the voltage and energy in a capacitor. Apparently you have based your entire thinking on MileHigh's hypothetical thought experiment that does not occur in reality.

Magluvin	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #217 on: 2025-03-07, 05:11:22 »
Group: Mad Scientist Hero Member Posts: 676	to say the least, milehigh was wrong. hypothetical? well at the time he was claiming it was so in order to counter my theory at the time. heck, every time i mentioned bifilar, he would attack. anyway.... im not saying there would be no heat via resistances. im not saying there would be no emmited radiations, all of which can be measured to equall that 50% loss. so why are those things not considered loss issues when we can get cap1 to cap2 in full minus the diode drop? no losses what so ever other than the diode drop? there is certainly resistance. there is certainly radiations. why are those not creating even small losses? and if they are instilling the losses we are talking about, then can we say cap1 through an inductor to cap2 is a form of excess energy, one that overcomes those losses to an exact, easily predictable outcome of seemingly no loss at all??? but that loss is not because of those things. and, those things used that transfer of charge the same way the inductor made use of the transfer. what i do see is just about everyone here firstly claimed that resistance was the cause. but then when i shut that down, then it must be radiations as a cause. so what is it exactly??? seemed so dead set on resistance because science says so. so, i suppose even science doesnt know either as they claim one or the other but no definite answers, yet the science is still the only authority even when proven wrong. at times i feel as im dealing with a cult. lol. if you dont believe things are being hidden from us in the science world, then you will find out one day. they lie about the vaccines. they lie about blood pressure and salt. they lie about co2. infact when climate, so called, experts were asked in congress what they all thought the co2 levels in the atmosphere are, and some had different answers. most ranged from 5% to even 15%. but the person asking the question took them all down, as the actual numbers were less than 1%. and not one of them had any rebuttal. not one. theres your truth. from scientists, experts, the ones we must trust. so those are only a speck in the sand of lies from science. why is it soo unbelievable that maybe im correct, when in actuality, you dont know for sure if im wrong. mags

web000x	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #218 on: 2025-03-07, 05:48:17 »
Sr. Member Posts: 340	Quote from: Magluvin on 2025-03-07, 05:11:22 to say the least, milehigh was wrong. hypothetical? well at the time he was claiming it was so in order to counter my theory at the time. heck, every time i mentioned bifilar, he would attack. anyway.... im not saying there would be no heat via resistances. im not saying there would be no emmited radiations, all of which can be measured to equall that 50% loss. so why are those things not considered loss issues when we can get cap1 to cap2 in full minus the diode drop? no losses what so ever other than the diode drop? there is certainly resistance. there is certainly radiations. why are those not creating even small losses? and if they are instilling the losses we are talking about, then can we say cap1 through an inductor to cap2 is a form of excess energy, one that overcomes those losses to an exact, easily predictable outcome of seemingly no loss at all??? but that loss is not because of those things. and, those things used that transfer of charge the same way the inductor made use of the transfer. what i do see is just about everyone here firstly claimed that resistance was the cause. but then when i shut that down, then it must be radiations as a cause. so what is it exactly??? seemed so dead set on resistance because science says so. so, i suppose even science doesnt know either as they claim one or the other but no definite answers, yet the science is still the only authority even when proven wrong. at times i feel as im dealing with a cult. lol. if you dont believe things are being hidden from us in the science world, then you will find out one day. they lie about the vaccines. they lie about blood pressure and salt. they lie about co2. infact when climate, so called, experts were asked in congress what they all thought the co2 levels in the atmosphere are, and some had different answers. most ranged from 5% to even 15%. but the person asking the question took them all down, as the actual numbers were less than 1%. and not one of them had any rebuttal. not one. theres your truth. from scientists, experts, the ones we must trust. so those are only a speck in the sand of lies from science. why is it soo unbelievable that maybe im correct, when in actuality, you dont know for sure if im wrong. mags Impedance is a complex quantity. It contains both resistance and inductive reactance. If the circuit impedance is primarily resistive, you will have most of the energy dissipate as heat losses. If the circuit is primarily inductive reactance, you will have stored magnetic energy that can then be transferred into C2. Dave

Magluvin	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #219 on: 2025-03-07, 05:59:57 »
Group: Mad Scientist Hero Member Posts: 676	one more thing... ive made analogies that i compare air pressure in tanks and water buckets. can you tell me what i am missing here? am i not correct in doing so even though the results are virtually the same in each case? voltage is pressure. current is flow due to pressure. results are the same. cap to cap, 10v to each having 5v. tank to tank, 100lb to 50lb each tank. 10gal to 2 buckets of 5 gal. please show me how i should not relate these as all the same. air tanks, transfer to tank2 with an air motor with a flywheel as the inductor. same with the water buckets, water pressure driven motor with flywheel as the inductor. can anyone agree with me on that? if so, in the air tank setup where we just open the valve and let the pressure equal out between tanks. 100lb to 50lb each. 50% loss. did we lose that 50% because of resistance or radiations??? if not then where did that energy go?? ill tell you where it went. we stupidly, pumped tank1 to 100lb and just dumped it into a container with twice the volume, which in the end adds up to a 50% loss. why is that so hard to understand when it comes to the cap to cap?? any resistance from air tank to air tank is not the cause of the loss in the end. maybe that air flow through the hose from tank to tank makes a bit of heat due to air friction. but it didnt cause the 50% loss, now tell me about how radiations caused the loss. please tell me how. you can calculate the amount of energy stored in the air tanks in the form of pressure by volume. you can quantify the amount of work that a 10gal bucket of water can do with the pressure from a hose outlet at the bottom of the bucket.. we can do work with that water flow caused by the weight of the water in the bucket. 10 gal in 1 bucket can do more work than 2 buckets with 5 gal, just like the caps from 10v in one cap to 2 at 5v. same exact thing. once you get it straght that they are all the same, then you will see that the resistance and radiations are not the cause of the loss. you then should also understand the the electron count on the caps plates that determine the voltage(pressure) are the same as the air molecule count in the tanks or the water molecule count in the buckets. dont assume AI is always correct. AI only has reference to info already out there that may be wrong intentionaly or not. dont be stuborn. really think about this. mags

verpies	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #220 on: 2025-03-07, 09:09:18 »
Group: Professor Hero Member Posts: 3670	Quote from: web000x on 2025-03-07, 05:48:17 Impedance is a complex quantity. It contains both resistance and inductive reactance. ...and capacitive reactance. Quote from: web000x on 2025-03-07, 05:48:17 If the circuit impedance is primarily resistive, you will have most of the energy dissipate as heat losses. If the circuit is primarily inductive reactance, you will have stored magnetic energy... ...and if the circuit contains primarily capacitive reactance, its energy will be stored mostly as electric field. Furthermore, the effects of inductive reactance can cancel the effects of capacitive reactance, ...but we don't know of anything that can cancel the effects of resistance. Especially its conversion of electric current into heat (@Mags: water pipes almost do not do that).

F6FLT	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #221 on: 2025-03-07, 11:23:41 »
Group: Experimentalist Hero Member Posts: 2148	Quote from: web000x on 2025-03-07, 05:48:17 ... If the circuit is primarily inductive reactance, you will have stored magnetic energy that can then be transferred into C2. Dave This is why charging one capacitor from another, through an inductor, can be done without loss (which is what happens in a resonant circuit). The inductor is charged by a current that smoothly charges the capacitor. It's as if we were charging with a voltage barely above the capacitor's voltage and gradually increasing it. This avoids the important voltage difference between the charging source and the voltage of the capacitor being charged, which in the ordinary case of a series resistor is compensated by it through energy loss. --------------------------- "Open your mind, but not like a trash bin"

Itsu	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #222 on: 2025-03-07, 11:35:21 »
Group: Elite Experimentalist Hero Member Posts: 4236	Quote from: verpies on 2025-03-06, 21:34:14 Just flip that cap around Of course, but people should know that we have a reversed polarity there, so it is not the same as in the normal cap to cap transfer circuit. After reversing C2, i still have the draining within 31 seconds. I added another switch between C2 positive and ground and now the draining stops, however during the opening of this switch we have a 140V spike as if we interrupt the magnetic field in a coil. Itsu ------------------------ Checking out Joel Lagace - Hang on I'M SERIOUS!! cap paradox plus inductor plus reversed plus 2nd switch small.png (61.62 kB, 952x427 - viewed 225 times.)

verpies	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #223 on: 2025-03-07, 15:14:18 »
Group: Professor Hero Member Posts: 3670	Quote from: Itsu on 2025-03-07, 11:35:21 After reversing C2, i still have the draining within 31 seconds. I added another switch between C2 positive and ground and now the draining stops, This means that the diode leaks. It is possible to make the CLC energy transfer without this diode by using a π circuit, but I find that circuit to be less educational. Quote from: Itsu on 2025-03-07, 11:35:21 I added another switch between C2 positive and ground and now the draining stops, however during the opening of this switch we have a 140V spike as if we interrupt the magnetic field in a coil. If there is current flowing in the coil when you open that switch, then you are opening it at the wrong time. Also LTSpice might be so smart that it simulates the diode's junction capacitance or even the DSRD effect.

Itsu	Re: Checking out Joel Lagace - Hang on I'M SERIOUS!! « Reply #224 on: 2025-03-07, 15:21:30 »
Group: Elite Experimentalist Hero Member Posts: 4236	Quote If there is current flowing in the coil when you open that switch, then you are opening it at the wrong time. Well, i made sure there is no current flowing in the inductor, see screenshot where SW2 is closed (150ms) when no current through L1 (i L1) is flowing, so IMO it can not be that. But I just noticed that even with SW2 open, we still have the drain on C2 where it is back to 0V after 5000 seconds (1.38 Hours). I guess its building time to see what is really happening..... Itsu ------------------------ Checking out Joel Lagace - Hang on I'M SERIOUS!! SW2 timing.png (59.05 kB, 1898x1032 - viewed 29 times.) Checking out Joel Lagace - Hang on I'M SERIOUS!! SW2 closed 1.38 hour later.png (18.59 kB, 1902x503 - viewed 31 times.)