Here's the clip:
Min2oly is explaining what "radiant energy" is to his audience. I am saying that he is making this clip without knowing what is really and truly going on. That's an unfortunate thing because if you do understand what is going on then your experimenting can be that much more interesting and rewarding.
There are some around here that have built one, two, perhaps even a dozen Bedini motors and have played with coils, capacitors, diodes and batteries in all sorts of different setups.
Anyone is free to contribute to the thread but the focus here is on the builders. If you build something you should know how it works. I am just going to watch from the sidelines and make some colour commentary. Try to keep your eye on the ball and discuss what is going on on the clip. When somebody really gets the explanation right I will shout, "Bingo!"
MileHigh
Magluvin is the first to contribute with these comments:
>>>>>>>>>>>>>>>>>>>>> MAGLUVIN FIRST COMMENT >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Its a boost converter.
When the battery is put across the coil, current begins to flow. As it flows, a magnetic field is built in the coil, and as the field builds, more current from the battery is taken.
When he disconnects the battery from the coil, the magnetic field collapses, causing currents to continue to flow through the coil, through the diode and into the cap, as it is the only short path available once the red wire is disconnected.. Mind you the current in the coil never changes direction. Wink
Didnt I do pages of this stuff at ou?
Tuff one.
Mags
>>>>>>>>>>>>>>>>>>> MAGLUVIN SECOND COMMENT >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Well how deep does the rabbit hole go?
More surface talk.....
Another possibility is, that I can only assume what orientation the diode is in, which determines the way current can flow through it.
Even if we reverse the diode, we still get our high voltage to the cap, but it is opposite polarity now.
When we charge a coil as shown in the vid, the field builds, and when we disconnect the source from the coil, that field collapses.
If we have a perfect switch used to connect and disconnect the source to the coil, meaning when on its on, but when off it will not allow current to flow, not even HV spark, then when we disconnect(open switch) the "forward" voltage of the collapse has only the path of the diode to charge the cap of my previous description of the vid.
What if we dont have a diode and cap? With our perfect switch, when we open it, the field collapses. And as it collapses down toward zero field, there is no place or path for the voltage potential produced by the collapsing coil to go.
Coils have capacitance. So the combination of coil with built in capacitance we have an LC. Single wire coils have very small capacitance between windings, and bifilar coils have a higher capacitance.
If you charge up an LC and disconnect, it will oscillate. So when we disconnect the source from our coil with no diode and no cap, our coil will oscillate at very high freq due to very small capacitance in our coil. The oscillation dies pretty quickly. No oscillation occurs if we have a path for that potential to the cap. The oscillation will be heavily damped
So if we reconnect our diode and cap, only we reverse the diode, when the source is disconnected from the coil, and the field collapses. But with no where for a continued forward potential current to flow, the coil goes into its oscillation mode, and the field collapses very quickly, and when the field polarity of the coil reverses during oscillation, N S to S N, that is when the potential of the coil has a path for current through the diode and to the cap for storage.
This is what I call back emf. In my original description, I call it forward emf, meaning that when we disconnect the source from the coil, the collapsing field current is in the same direction as the current produced by the source through the coil.
In this reversed diode description, once the field of the coil reverses within 1/4 wave of LC oscillation, then the output of the coils current is in the opposite direction of the original source current. Back emf.
The output to the cap in either situation will be nearly the same. Possibly a small, tiny, itiy bitty, difference due to the bemf version may have lost some due to the dying oscillation during that 1/4 wave of oscillation mode, being that the oscillation eventually dies. So there is some loss in each cycle.
There are some out there that use the term bemf when actually to be correct in some situations, should use just emf or forward emf, produced by the collapsing field of a coil.
Well, im beat. Dont really have time to edit errors.
Magsleeps soon
Magluvin:
QuoteEven if we reverse the diode, we still get our high voltage to the cap, but it is opposite polarity now.
That's not correct. Sometimes it helps to make a simple drawing on paper. You might want to try that and see what comes up.
For your discussion about a open-circuit coil ringing because of the self-capacitance and all that, the essence of what you were saying was right but it has nothing to do with the clip.
Again, bringing it back to the clip, for anyone and especially the builders: What is really happening in Minn2oly's clip?
MileHigh
Anytime a pulse motor builder gives me flack about "not being a builder" I am going to point them to this thread.
Again: Explain what is really going on in this clip and point out where Min2oly is wrong.
My take on it is that there is no point in building pulse motors if you can't dissect this clip and understand exactly what is going on.....
Revenge of the nerds!
MileHigh
Hey,
I just watched that clip for the first time. All I can say is:
OMG! C.C
There probably is some radiant energy produced when the handheld jumper loses contact with the diode inductor junction - most likely it is present in the RF bands where it could be easily detected with a nearby AM radio (since radio energy is radiant energy).
However, this experiment is demonstrating the capture of the BEMF spike produced at that moment the jumper loses contact.
1. Jumper makes contact and current immediately flows through the 8-filar series connected inductor but is temporarily limited by the inductive reactance.
2. A magnetic field forms around the inductor commensurate with current flow which increases with time.
3. Jumper loses contact and current flow is abruptly ended
4. Magnetic field collapses into the inductor and attempts to keep the current flowing but the path is broken
5. Charges accumulate across the coil and are measured at voltage. Normally these charges would 'ring' down until dissipated but the diode prevents that.
6 Voltage present at the cathode of the diode is retained thus charging the capacitor.
7. The meter gives us an approximated integration of the capacitor charge - the meter also loads the circuit with some internal path thus slowly discharging the capacitor.
What would be more interesting here - is a discussion on the best way to maximize the efficiency of the process - Joule Ringer comes to mind http://www.youtube.com/watch?v=glRuwV9IlaY (http://www.youtube.com/watch?v=glRuwV9IlaY)
Quote from: MileHigh
...
Again: Explain what is really going on in this clip and point out where Min2oly is wrong.
...
It is possible to perform that task in
a positive, constructive and educationally
sound fashion.
Pretend that you are Min2Oly.
How would you desire someone to step
forward and show you the error?
Could you provide us an example of how
this might best take place?
Thanks for the go at it Harvey but still looking for the answer. Hoping a pulse motor builder will post.
Dumped, just chill with the politically correct social engineering please, it's oppressive.
@Poynt99
QuoteI just watched that clip for the first time. All I can say is:
OMG!
I just watched the video and I would agree, Double OMG, I can almost feel the radiant energy coming off that puppy from all the way up here in Canada, the man is a musical genius just like Stevie Wonder. On another note something is whacked in that video, either his diode is pooched the meter is shot or he is using an electrolytic cap connected backwards because there is no way in hell the cap should bleed off that fast.
Regards
AC
Quote from: MileHigh
...
Dumped, just chill with the politically correct social engineering please, it's oppressive.
Just an appeal to bring out the civility and
breeding that resides within you.
Nay, not social engineering - but I can see
where one might think so.
That you desire to cling to a manner of behavior
which you find mischievously delightful in defiance
to encouragement to act otherwise may indeed
seem "oppressive." It may be that you perceive
it to be your principal means of wielding "power"
over your peers - thus, to give it up would be a
most difficult decision.
The art of being a gentleman seems long lost.
Dumped,
I think you should try getting into a contest with AC, the two of you might be like pees in a pod. I am not going to engage you, the 9/11 debate was a life's worth for me.
MileHigh
Hello MileHigh, My first post here. I think at about 4:20 mins in the video he gets a bit mixed up and says the radiant spike gets to the capacitor before the current flows through the short back to the battery. I think what is actually happening is the short is charging the coil and building a magnetic field, when the circuit is opened the magnetic field collapses and charges the cap. First he applies the voltage then current charges the coil, the circuit is then opened and the magnetic field then collapses producing emf to charge the cap, there is current involved in that too.
There is still current flowing through the short but if the connection is brief enough maximum current won't happen. I think his coil has too much resistance, the cap is too small and probably should be an electrolytic one.
I haven't built many pulse motors though just two.
Cheers
@Farmhand
QuoteThere is still current flowing through the short but if the connection is brief enough maximum current won't happen. I think his coil has too much resistance, the cap is too small and probably should be an electrolytic one.
I would not use the term "short" but rather "path", there are two paths 1)from the battery through the coil when the jumper closes the circuit and 2)from the battery through the coil, the diode, the capacitor and back to the battery when the jumper opens path 1. It should be noted the battery is in series with the discharging coil and that the coil acts as a current source when discharging. It should also be noted that the converter becomes very inefficient once the magnetic field of the coil is no longer expanding yet current is still flowing, in fact this energy is completely wasted relative to the charging of the capacitor.
Regards
AC
G'day AC, Yeah path is maybe a better term, I was thinking along the lines of the circuit is already complete but an idle/open circuit (because of the cap) until he taps the negative of the coil to the negative of the cap/battery thereby completing the path (closed circuit) from the battery through the coil. :) The second path you indicated is not made until he manually connects the wire and is not how you say it is if I am seeing the video correctly. He put's the capacitor in parallel with the battery after the cap is charged. When he discharges the capacitor into the battery the coil and diode are not in the active circuit path. From what I can tell. I drew a picture of the circuit and I'm looking at it now, your right, the diode prevents shorting of the cap but closes the circuit by bypassing the cap and diode.
That's how it looks to me.
Also. The way I see it the coil produces emf when the magnetic field collapses, and it's that emf which causes the current which in turn I think induces back emf. Although the back emf is pretty much irrelevant to this conversation in my opinion. But it is worth mentioning I guess.
My main points would be that the emf from the battery when applied across the coil causes current flow, this charges the coil (builds the magnetic field), then when the circuit is opened and the magnetic field collapses the coil becomes a source of emf, which causes current flow if the circuit is configured to allow it.
Cheers
Welcome to the forum Farmhand. I have seen your comments on OU and on YouTube and they show that you have your feet firmly planted on the ground.
Lots of you have made good comments mixed in with some bad comments and you are starting to converge on the solution but still no Bingo! Farmhand you are right that Min2oly's comment at 4:20 is wrong. It's pure "old wives' tale" material and let's say it's the equivalent of speaking English and not being able to conjugate the verbs "to have" and "to be" properly. It's a grave error he is making.
As a hint, it also makes sense to look at examples like this in terms of energy also. i.e.; Discuss what is happening in the clip when you look at the energy.
Also, no one has discussed the core vs. no core aspect.
Still challenging any pulse motor builders besides Farmhand to chime in.
MileHigh
"...Still challenging any pulse motor builders..."
Is this a pulse motor?
http://www.youtube.com/watch?feature=player_detailpage&v=m6DiE81UBfs (http://www.youtube.com/watch?feature=player_detailpage&v=m6DiE81UBfs)
8)
Is this a pulse motor? Hehe hey Harvey! How ya doin?
Here is a pulse motor, using the back emf, or what ever one might call it at any time, place or way, to a 5 ohm 5w resistor.
http://www.youtube.com/watch?v=b7PR8JEVp7A
Mags
Mags and Harvey:
So you guys get your hands dirty, good for both of you. (Whoops! I think Farmhand mentioned that he is a builder too.)
From my experience online, the problem is that perhaps 9 out of 10 pulse motor builders could not even begin to comment on Min2oly's clip. Perhaps 4 out of those 10 also get "corrupted" like Min2oly.
Mags I watched your clip and I have an observation and a related question. The observation is that most pulse motor builders either don't try to measure the average output power from the inductive spike that goes to the charging battery or they believe that it's impossible to measure the power of the "radiant energy."
In your clip from 2009, you have a 5-watt resistor that gets quite hot. You don't have a charging battery and instead you just have the 5-watt resistor. So the question is have you ever tried to measure the average output power of the inductive spike?
Just curious if the build is still intact and sitting on a shelf somewhere.
MileHigh
I dont have that setup at this time. I have all of the coils rotor and mags etc. It isnt too difficult to setup.
At the time I believed I was showing the average power of the spike, by routing it to a cap via a diode then to the 5ohm resistor.
Tell me more.
Mags
MileHigh,
Do you think it is possible to use a very short (in the nanoseconds range) pulse onto a wire
or a coil and then extract the first short voltage spike before any real current is starting to
flow in that wire or coil?
GL.
Mags and Groundloop:
If you are going to verbally describe a simple circuit please give a complete description. I will try to comment anyways.
Mags, if you have your cap in parallel with the 5-ohm resistor and they are connected to ground, and the coil output is feeding the cap/resistor on the other side, chances are that the cap is smoothing out the pulsing to a nice DC value. That's how you measure the average power out from the pulsing coil. You didn't say the value of the cap though. Also, if you can measure the average output power, you are well on the way to measuring the value of the inductance of the coil.
Also, do you have a scope?
So if we assume that the voltage across the cap is near-DC and you are measuring 2.5 volts that would be 1.25 watts dissipated in the resistor that wouldn't make sense because you said the resistor was running very hot. So chances are something else was going on.
Notice that you have a very low resistance of only five ohms. Who says that you can lower and lower the resistance without some other effect affecting the average output power from the coil? So that's the big hint, can you think about what may coming into play there? Also, if you are going to split hairs, you should measure the output power with the voltage across the cap at a stable 12.6 volts (or thereabouts) to emulate the missing charging battery. Another brain teaser, why would that be the case?
For Groundloop, when you say extract the short voltage spike, I assume that you mean the energy in the spike. Is that correct?
We know that EM waves travel about 30 cm in empty space in one nanosecond. Let's say that a voltage signal travels about 10 cm through wire in one nanosecond.
Suppose your coil consists of a length of wire that's 1000 cm long (10 meters). So if you make a 10 nanosecond pulse that will travel along the wire and be about 100 cm in length. So in this case the coil is not even a coil with the property of inductance anymore. It acts like a microwave transmission line in this case. Once frequencies become very high microwave transmission line effects take over. You can still work out exactly what happens but you are in a different realm that is above and beyond basic electronics.
Again, for both of you, you must describe exactly what the configuration of your circuit is if you want to understand it. Electronics is a precise science and talking in generalities without describing things properly really doesn't work. I may sound anal but it's the truth.
MileHigh
Quote from: MileHigh on 2012.02.15, 02:16:02
Mags and Groundloop:
If you are going to verbally describe a simple circuit please give a complete description. I will try to comment anyways.
Mags, if you have your cap in parallel with the 5-ohm resistor and they are connected to ground, and the coil output is feeding the cap/resistor on the other side, chances are that the cap is smoothing out the pulsing to a nice DC value. That's how you measure the average power out from the pulsing coil. You didn't say the value of the cap though. Also, if you can measure the average output power, you are well on the way to measuring the value of the inductance of the coil.
Also, do you have a scope?
So if we assume that the voltage across the cap is near-DC and you are measuring 2.5 volts that would be 1.25 watts dissipated in the resistor that wouldn't make sense because you said the resistor was running very hot. So chances are something else was going on.
Notice that you have a very low resistance of only five ohms. Who says that you can lower and lower the resistance without some other effect affecting the average output power from the coil? So that's the big hint, can you think about what may coming into play there? Also, if you are going to split hairs, you should measure the output power with the voltage across the cap at a stable 12.6 volts (or thereabouts) to emulate the missing charging battery. Another brain teaser, why would that be the case?
For Groundloop, when you say extract the short voltage spike, I assume that you mean the energy in the spike. Is that correct?
We know that EM waves travel about 30 cm in empty space in one nanosecond. Let's say that a voltage signal travels about 10 cm through wire in one nanosecond.
Suppose your coil consists of a length of wire that's 1000 cm long (10 meters). So if you make a 10 nanosecond pulse that will travel along the wire and be about 100 cm in length. So in this case the coil is not even a coil with the property of inductance anymore. It acts like a microwave transmission line in this case. Once frequencies become very high microwave transmission line effects take over. You can still work out exactly what happens but you are in a different realm that is above and beyond basic electronics.
Again, for both of you, you must describe exactly what the configuration of your circuit is if you want to understand it. Electronics is a precise science and talking in generalities without describing things properly really doesn't work. I may sound anal but it's the truth.
MileHigh
MileHigh,
Let me start by saying that I know that electronics is a precise science, and I also know what a transmission line is.
And it is not a problem for me to design and build a switch that can switch on and off in the nanoseconds range.
>>For Groundloop, when you say extract the short voltage spike, I assume that you mean the energy in the spike. Is that correct?
No, just a voltage potential and no current transferred over the transmission line.
And you did not answer my question. So I will ask again:
Do YOU (as a person) think it will be possible to put a short nanosecond pulse onto a transmission line of some length,
then switch off the input pulse before any current has started to flow over the transmission line, and then switch on
a load at the other end of the transmission line to capture energy into a load?
A simple yes or no answer will do fine.
GL.
Quote from: Magluvin on 2012.02.14, 02:47:42
Is this a pulse motor? Hehe hey Harvey! How ya doin?
Here is a pulse motor, using the back emf, or what ever one might call it at any time, place or way, to a 5 ohm 5w resistor.
http://www.youtube.com/watch?v=b7PR8JEVp7A
Mags
Hiya Mags,
Doin ok I suppose - could use about 40K to dig myself out of this economic rabbit hole I fell into though :P
Haven't been online much lately but the forum here sent me an email saying it was going to drop my account so I figured I should probably pop in and say hi at least O0
What ever happened with the mini MEG . . . ORBO BonBon thing - did you ever get any surplus out of that config? I was going to analyze Bearden's take on all that but first I wanted to dig into Heaviside's stuff since Bearden was throwing his name in the mix and when I started on that I realized I needed to read Maxwell's original paper - and about half way through that I got sidetracked and never finished. Sure would be nice to pin down exactly what energy pool all these fellows are extracting from, i.e. T.H. Moray, Tesla, Marks, Schwartz etc. Perhaps Dirac was right and there is something there to balance it all out. Interesting thing about Schwartz' work is that energy output would be slightly higher facing the sun than facing away and definitely showed extreme variance related to distance from the equator.
On another note there should be a lot more sporadic claims coming in as we approach the solar maximum and the sun spots wreak havoc on all the testing going on globally.
I hope Milehigh gets the answer he's looking for. I'm pretty sure my analysis is spot on, but I'm too apathetic to care either way :P
Best wishes to all.
8)
Harvey:
There is a possibility that your analysis is spot on but your write up is sometimes ambiguous. When I first read it I may have misconstrued ambiguity in your prose as flat-out errors. Describing a process is tricky.
Groundloop:
QuoteDo YOU (as a person) think it will be possible to put a short nanosecond pulse onto a transmission line of some length,
then switch off the input pulse before any current has started to flow over the transmission line, and then switch on
a load at the other end of the transmission line to capture energy into a load?
No. Power flows through the transmission line, the wave has to have an electric and a magnetic component. The rate of change of the electric field with respect to time will induce a corresponding magnetic field as determined by the impedance of the transmission line.
In a related matter, there is no such thing as "one wire energy transmission." All of those demos rely on high-frequency AC and capacitive coupling always provides the "invisible second wire."
MileHigh
Quote from: MileHigh on 2012.02.16, 00:16:04
Harvey:
There is a possibility that your analysis is spot on but your write up is sometimes ambiguous. When I first read it I may have misconstrued ambiguity in your prose as flat-out errors. Describing a process is tricky.
Groundloop:
No. Power flows through the transmission line, the wave has to have an electric and a magnetic component. The rate of change of the electric field with respect to time will induce a corresponding magnetic field as determined by the impedance of the transmission line.
In a related matter, there is no such thing as "one wire energy transmission." All of those demos rely on high-frequency AC and capacitive coupling always provides the "invisible second wire."
MileHigh
MileHigh,
Thank you for answer my question. So, no hope then, just give up? :-)
I think your right on the second one, but not 100% right on the first one. If we apply only a voltage potential onto the line
and cut of the potential before any real current can flow, then I think nature will provide us with the magnetic component
and the current that creates it. Nature will always try to fill a void. If we use only a ultra short voltage spike then there will
be a lack of current and a magnetic field and nature will see that as a void. So when the pulse has traveled to the other
end of the line then there will be a current and magnetic component that gives us real power. That is the idea anyway.
Comment?
GL.
Quote from: Groundloop on 2012.02.16, 01:29:05
If we apply only a voltage potential onto the line and cut of the potential before any real current can flow...
GL & MH,
I hope you folks don't mind a comment from me. This question and the above statement fit into some of my favorite past research.
If you apply a potential and disconnect it before current can flow then that potential will not be on the wire. Current must flow for the potential to appear on the wire.<Unless we are speaking of electrostatic induction.>
If these thoughts are related to the story of Tesla looking into people being killed due to power being switched on a cable, then I think the thought has gone astray. I doubt it was ever a matter of opening the circuit before current could flow. I think it was making the time for the open cycle so short that the potential fall or trailing edge would fit within the length of wire.
When this is done you can have the most vigorous mechanical movement of unsecured wire. This may have been misconstrued as the effect of some special wave traveling down the wire in Tesla's day.
Quote from: Groundloop on 2012.02.16, 01:29:05
If we apply only a voltage potential onto the line and cut of the potential before any real current can flow...
If we use only a ultra short voltage spike then there will be a lack of current and a magnetic field...
So when the pulse has traveled to the other end of the line then there will be a current and magnetic component that gives us real power.
GL,
The instant you apply potential to a circuit or even an open transmission line (TL), current begins. Remember you are charging a series L, and shunt C in the TL.
Keep in mind the relaxation time of an electron (http://www.answers.com/topic/relaxation-time-of-electrons). You would need to switch ON and OFF your potential in much less that 10
-14 seconds (100 fempto-seconds) before having any hope of disconnecting before any conduction current occurs.
.99
Quote from: WaveWatcher on 2012.02.16, 02:00:26
GL & MH,
I hope you folks don't mind a comment from me. This question and the above statement fit into some of my favorite past research.
If you apply a potential and disconnect it before current can flow then that potential will not be on the wire. Current must flow for the potential to appear on the wire.<Unless we are speaking of electrostatic induction.>
If these thoughts are related to the story of Tesla looking into people being killed due to power being switched on a cable, then I think the thought has gone astray. I doubt it was ever a matter of opening the circuit before current could flow. I think it was making the time for the open cycle so short that the potential fall or trailing edge would fit within the length of wire.
When this is done you can have the most vigorous mechanical movement of unsecured wire. This may have been misconstrued as the effect of some special wave traveling down the wire in Tesla's day.
WW,
>>Current must flow for the potential to appear on the wire.
I agree on that one but I think that nature will provide that current because nature will not allow a void to exists.
To harvest free energy all you have to do is to create a void. Nature will always fill that void for you. The problem
is to actually create that void without using any real power.
GL.
Quote from: poynt99 on 2012.02.16, 02:06:45
GL,
The instant you apply potential to a circuit or even an open transmission line (TL), current begins. Remember you are charging a series L, and shunt C in the TL.
Keep in mind the relaxation time of an electron (http://www.answers.com/topic/relaxation-time-of-electrons). You would need to switch ON and OFF your potential in much less that 10-14 seconds (100 fempto-seconds) before having any hope of disconnecting before any conduction current occurs.
.99
.99,
I never said that it would be easy. :-)
GL.
GL,
Are you thinking that there is momentum in the application of a potential? I think that is true but by then current is flowing and any conventional thoughts of momentum can no longer be applied.
So, this would be similar to striking the end of a very long rod with a single hammer blow. Such a 'wave' would be akin to a soliton or compression wave. A very hard one to demonstrate, I think.
The question is:
Even if you were able to disconnect before those first few electrons collide with the next, haven't you still imparted some energy into those first few electrons? I think the answer to that is YES. And as such, haven't you imparted that energy from your source of potential?
.99
I must agree that no matter how short the pulse, the energy circulate that wire would comes from your input.
Now let's say we have an LC circuit and injected some energy into it. Let's also make this LC circuit superconducting. Will the energy flow back and forth forever?
Provided both the L and C are ideal, and there is nothing else connected, YES.
@Groundloop
QuoteIf we apply only a voltage potential onto the line
and cut of the potential before any real current can flow, then I think nature will provide us with the magnetic component
and the current that creates it. Nature will always try to fill a void. If we use only a ultra short voltage spike then there will
be a lack of current and a magnetic field and nature will see that as a void. So when the pulse has traveled to the other
end of the line then there will be a current and magnetic component that gives us real power. That is the idea anyway.
I remember one time back on the farm we had a 1/2 mile stretch of 2" ABS pipe from a pump in the dugout and this pipe went to the yard where there was a ball valve with a pressure guage. Now if I opened and then closed the ball valve really fast the pressure would fall below the 50psi standard pressure then spike to 100psi then fall to almost zero and this cyclic wave like process would slowly be dampened over time until the pressure was back to the original 50psi. I understand some might just say hmmm and just walk away however I was never that way and whenever I see something peculiar I am compelled to experiment until I fully understand the phenomena. As it turns out opening the valve and letting ten gallons of water spill on the ground is no more effective at producing the pressure waves than one ounce because the phenomena is not dictated by the magnetude of flow persay but how fast the valve closes. The faster the valve slams shut the greater the pressure developed and the longer the wave like motion persists due to this higher pressure.
Next there is a condition whereby one can "pump" the wave however it is not what one might expect, the most effective way to parametrically pump the sytem is when the pressure is the lowest at the guage. Now consider the mechanics here, we have a fluid in a pipe and a wave of pressure is travelling back and forth from one end to the other. When the pressure is highest at one end it may be at it's lowest at the other end, that is one end is above ambient pressure and the other below ambient pressure. To me it would seem rather silly to open the valve at the highest pressure because the pressure is what is responsible for the motion of the wave. Now if one opens the valve at the lowest pressure a given volume of water has been removed from the system making the already low pressure lower thus the difference between high and low is greater not less. As it turns out the most efficient way to maintain a high pressure wave involves the least amount of water leaving the system, go figure,and it also involves the least amount of pressure.
When I look back through the years it was always the subtle things from which I gained the greatest understanding, those peculiar little things which normally go unnoticed by the other 99% of people. For instance one time I had an inflatable pool 4' deep and 15' in diameter and when I pushed slowly on the side the familiar standing waves would emerge on the surface. I found it curious that with a circle one could push anywhere on the side and the waves would emerge from every edge of the pool simultaneously. However after many experiments regarding the frequency, magnitude and rate it was found that a quick hard slap on the side would produce a wave at the very center of the pool which was 10 times higher than any of the ordinary "ripples". A virtual geyser of water streaming upward sometimes as high as two feet and four inches in diameter from one single impulse. I have found many people have a good understanding of many things however many times it is the level of understanding which matters the most, the subtle little things which normally go unnoticed which can lead to a new way of looking at something, perspective matters
Regards
AC
Here's a very short clip I shot. It shows when I first energize the circuit the neon connected
between the drain of one mosfet and circuit ground flashes, also mostly when turned off as well.
It appears as though the leg of the neon connected to circuit ground is lighting only.
The supply is 12 volts and the circuit is switching to make AC at 90 Hz.
http://www.youtube.com/watch?v=CnJE_K5_XSU
Being that I am very new to all this, I have some questions. I hope MileHigh doesn't mind me asking
them here, I think it's kinda related to the subject at hand.
Q1. How would i accurately describe what I'm seeing there ?
Q2. What is the cause of it ?
Q3. How do I stop it ?
It annoys me. >:( If there is a problem with the setup at times it can cause the neon to flash even
during normal operation and that is an indication the mosfets will be heating up quick.
Q4. Could it be caused by too much dead time? EDIT: (Causing it to happen during normal operation at certain times I mean.)
Oh and Q5. How come it only ever happens with that mosfet and not the other one ? Even if I switch the wires around. :-\
Cheers
Quote from: allcanadian on 2012.02.16, 05:43:12
@Groundloop
I remember one time back on the farm we had a 1/2 mile stretch of 2" ABS pipe from a pump in the dugout and this pipe went to the yard where there was a ball valve with a pressure guage. Now if I opened and then closed the ball valve really fast the pressure would fall below the 50psi standard pressure then spike to 100psi then fall to almost zero and this cyclic wave like process would slowly be dampened over time until the pressure was back to the original 50psi. I understand some might just say hmmm and just walk away however I was never that way and whenever I see something peculiar I am compelled to experiment until I fully understand the phenomena. As it turns out opening the valve and letting ten gallons of water spill on the ground is no more effective at producing the pressure waves than one ounce because the phenomena is not dictated by the magnetude of flow persay but how fast the valve closes. The faster the valve slams shut the greater the pressure developed and the longer the wave like motion persists due to this higher pressure.
Next there is a condition whereby one can "pump" the wave however it is not what one might expect, the most effective way to parametrically pump the sytem is when the pressure is the lowest at the guage. Now consider the mechanics here, we have a fluid in a pipe and a wave of pressure is travelling back and forth from one end to the other. When the pressure is highest at one end it may be at it's lowest at the other end, that is one end is above ambient pressure and the other below ambient pressure. To me it would seem rather silly to open the valve at the highest pressure because the pressure is what is responsible for the motion of the wave. Now if one opens the valve at the lowest pressure a given volume of water has been removed from the system making the already low pressure lower thus the difference between high and low is greater not less. As it turns out the most efficient way to maintain a high pressure wave involves the least amount of water leaving the system, go figure,and it also involves the least amount of pressure.
When I look back through the years it was always the subtle things from which I gained the greatest understanding, those peculiar little things which normally go unnoticed by the other 99% of people. For instance one time I had an inflatable pool 4' deep and 15' in diameter and when I pushed slowly on the side the familiar standing waves would emerge on the surface. I found it curious that with a circle one could push anywhere on the side and the waves would emerge from every edge of the pool simultaneously. However after many experiments regarding the frequency, magnitude and rate it was found that a quick hard slap on the side would produce a wave at the very center of the pool which was 10 times higher than any of the ordinary "ripples". A virtual geyser of water streaming upward sometimes as high as two feet and four inches in diameter from one single impulse. I have found many people have a good understanding of many things however many times it is the level of understanding which matters the most, the subtle little things which normally go unnoticed which can lead to a new way of looking at something, perspective matters
Regards
AC
AC,
There are a couple of strange things I have observed with electronic circuits. Look at the attached circuit. It is a normal boost
converter I did made for charging lead acid batteries. When I did test this circuit I did by mistake put a depleted lead acid battery
at the input (left in circuit) and a almost fully charged battery on the output (right in circuit drawing). The depleted lead acid battery
was at a very low voltage (approximately 6 volt). The circuit oscillator did run and one should think that the circuit should have
drained the input battery even more and charged the output battery up a little bit more. But that was not what happened. The
input battery did start to charge and the output battery did start to discharge. So the energy in the circuit was traveling the
wrong way of what is logical. Anybody up to the task of explaining why that did happen?
GL.
Quote from: Groundloop on 2012.02.16, 09:42:03
Anybody up to the task of explaining why that did happen?
An example of a rare but known action....
The key is the output transformer. When the oscillator runs 1 half of the cycle the transformer acts as a magnetic amplifier. This causes energy to flow from the output battery to the input battery, when the input battery voltage is low and the transformer is ran near saturation.
You can prevent the problem by using a physically larger transformer and sometimes by reversing the output circuit on the transformer output leads.
I've seen this issue claimed as verification of aether flow and Bedini's inventions :)
GL,
I think this is a good example of nature abhorring a vacuum ;)
BTW: Anytime someone with a 3(or more) coil JT starts to claim weirdness - the above is most probably the reason.
Quote from: poynt99 on 2012.02.16, 04:49:02
Provided both the L and C are ideal, and there is nothing else connected, YES.
How can we make the L and C ideal?
Quote from: GibbsHelmholtz on 2012.02.16, 13:34:44
How can we make the L and C ideal?
With present technology, we can't.
So how can we make it ideal...
@GibbsHelmholtz
QuoteSo how can we make it ideal...
We have had sufficient technology to do this for decades however what most lack is understanding, research the Leedskalnin PMH, ---- http://www.leedskalnin.com/LeedskalninsPerpetualMotionHolder.html. Why is it the experts here cannot maintain a current nor oscillation for more than a second or so while the student at this site has measured a current for weeks on end?. As Ampere once stated --- "It costs nothing to maintain a field only to create it", so why is it none of the experts here can seem to maintain a field for more than a second or so?. I did research on this subject and built some circuits and found that in fact what most call efficient is anything but which unfortunately was no real surprise.
Regards
AC
Quote from: allcanadian on 2012.02.16, 16:03:02
@GibbsHelmholtz We have had sufficient technology to do this for decades however what most lack is understanding, research the Leedskalnin PMH, ---- http://www.leedskalnin.com/LeedskalninsPerpetualMotionHolder.html. Why is it the experts here cannot maintain a current nor oscillation for more than a second or so while the student at this site has measured a current for weeks on end?. As Ampere once stated --- "It costs nothing to maintain a field only to create it", so why is it none of the experts here can seem to maintain a field for more than a second or so?. I did research on this subject and built some circuits and found that in fact what most call efficient is anything but which unfortunately was no real surprise.
Regards
AC
Hey Thanks AC,
It's interesting that you can store energy like that. It almost like current store in a coil and when disrupt it, the field collapsed. How can it runs in there without loss to resistance though. We can store energy in a cap, but there is no resistance in this case. I see the PMH current go in one direction, wonder if we can make it go back and forth.
Quote from: allcanadian on 2012.02.16, 16:03:02
@GibbsHelmholtz We have had sufficient technology to do this for decades however what most lack is understanding, research the Leedskalnin PMH, ---- http://www.leedskalnin.com/LeedskalninsPerpetualMotionHolder.html. Why is it the experts here cannot maintain a current nor oscillation for more than a second or so while the student at this site has measured a current for weeks on end?. As Ampere once stated --- "It costs nothing to maintain a field only to create it", so why is it none of the experts here can seem to maintain a field for more than a second or so?. I did research on this subject and built some circuits and found that in fact what most call efficient is anything but which unfortunately was no real surprise.
Regards
AC
Please post a link or a direction to where the current was measured continuously.
The last thing I read on it was his current meter needle was simply deflected by the magnetized metal. If that was wrong then something is up. If things are the way I researched, all of it is explainable with conventional information, albeit still not accepted by some academics. Usually, the types that Prof. Lewin mentioned as unable to really understand the basics of Faraday over Kirchoff.
I think my question is this:
If we have an LC circuit with energy flowing. Cap and coil resistance are zeros. How can we extract that energy without touching the circuit.
@GibbsHelmholtz
QuoteIt's interesting that you can store energy like that. It almost like current store in a coil and when disrupt it, the field collapsed. How can it runs in there without loss to resistance though. We can store energy in a cap, but there is no resistance in this case. I see the PMH current go in one direction, wonder if we can make it go back and forth.
That is an interesting question isn't it, I can take a pendulum on magnetic bearings in a vacuum and store energy as oscillations indefinitely, I can take a capacitor and charge it and store energy indefinitely however when I try to store energy as a static unchanging field in a coil of wire I must continually input energy to maintain this field despite Ampere's statement ---"It costs nothing to maintain a field only to create it". To prove this matter for myself I built a small electromechanical circuit and found that in fact the LC circuit most have assumed as being efficient is anything but and is a poor way of doing things. My electromechanical circuit could maintain it's oscillations and store energy hundreds of times longer than any equivalent LC circuit, which relates to my statement in regards to understanding and expertise. A person who never questions anything or seeks ways to improve it is no expert in my opinion.
Regards
AC
Farmhand:
Your questions can't be answered. As a bare minimum you need a schematic and a description of what you think your circuit is doing. Preferably you would include scope shots.
Groundloop:
QuoteThe depleted lead acid battery
was at a very low voltage (approximately 6 volt). The circuit oscillator did run and one should think that the circuit should have
drained the input battery even more and charged the output battery up a little bit more. But that was not what happened. The
input battery did start to charge and the output battery did start to discharge.
You are making the same mistake that I have just about given up on trying to educate people on. If you have a 12-volt battery that's outputting 6 volts it's "beyond dead," it's very very sick. Did you try measuring its output impedance? Then when you see the voltage increase from 6 to 7 volts even through you know it is driving a load and discharging you say the battery is "charging." That's nonsense, you know the battery is discharging. The zombie battery voltage is starting to go up simply because current is flowing through it and chemical reactions are starting to take place and some internal heat is being generated. Heat speeds up chemical reactions and voila you see more voltage at the terminals.
This idea that battery voltage is a reliable indicator of the state of charge of a battery is one of the blights that infects the free energy community like a cancer. I wish that it could simply be cut out. That's my 300th battery voltage rant.
I must have told poor soul Rosemary Ainslie that at least 10 times and she ignored it every single time.
MileHigh
Quote from: GibbsHelmholtz on 2012.02.16, 17:43:03
I think my question is this:
If we have an LC circuit with energy flowing. Cap and coil resistance are zeros. How can we extract that energy without touching the circuit.
Gibbs, read this pdf. Its not long, but it can answer your question as for one way of doing it. ;]
Mags
WaveWatcher:
QuoteThe key is the output transformer. When the oscillator runs 1 half of the cycle the transformer acts as a magnetic amplifier. This causes energy to flow from the output battery to the input battery, when the input battery voltage is low and the transformer is ran near saturation.
I don't know what's going on in your treatise. There is a diode restricting the direction of the current flow on the output battery so I don't see any mechanism for the energy to flow from the output battery to the input battery.
AC:
From your link:
QuoteI originally began this project to demonstrate the possibility of perpetual motion while studying at Purdue University in May of 2002.
Researchers have continued to be interested in this device because it defies natural laws of physics and is unexplainable by conventionally excepted electromagnetic theory. Modern science says it is impossible, but Edward Leedsklanin's theory says it is simple, natural, and that it's workings are misunderstood because there are some important aspects of modern science being taught in schools that are wrong.
Perhaps more intriguing than the ability to store energy indefinitely, the Perpetual Motion Holder can act like a living thing; and because it operates on the principles of a theory with a sound base of understanding, Leedskalnin is able to use this device as a real working depiction of the atom and the earth. Throughout his writings Leedskalnin points to a lack in a base understanding in all branches of science, and is quite thorough in his disparaging of J.J. Thomson's electron invention.
I know that this has been dealt with in the thread already but I have to chime in. Are the words above the writings of Leedskalnin himself? How can you possibly fall for that clap-trap AC?
This is an example of how people can be led astray just like Min2oly has been led astray by the Bediniites.
Everything that is highlighted above is either utter nonsense or an absolute lie. What is clearly shown in the prose above is that the person that wrote it has no clue, doesn't know his ass from his elbow when it comes to electricity and magnetism. I have heard this name Leedskalnin many times. This is this type of person that people in the free energy community are looking up to?
This is nothing more than a parlor trick done with magnetism. I doubt it's ever even discussed in classes. It sure as hell wasn't discussed when I was in school.
I am shaking my head.
MileHigh
AC:
QuoteI can take a capacitor and charge it and store energy indefinitely however when I try to store energy as a static unchanging field in a coil of wire I must continually input energy to maintain this field despite Ampere's statement ---"It costs nothing to maintain a field only to create it".
Gibbs:
QuoteI think my question is this:
If we have an LC circuit with energy flowing. Cap and coil resistance are zeros. How can we extract that energy without touching the circuit.
Put a loop of wire near the LC tank circuit so that it cuts through some of the alternating magnetic flux. Connect a load resistor across the two ends of the loop of wire. Without touching the LC tank circuit you will extract energy from it. When you do this the voltage and current in the LC tank circuit will decrease over time.
MileHigh
Quote from: Magluvin on 2012.02.16, 23:49:33
Gibbs, read this pdf. Its not long, but it can answer your question as for one way of doing it. ;]
Mags
Wow Mag,
Powerful stuff. Variant one on the pdf fitted the requirement to draw energy without touching (just by induction). What I think the reason variant one draw more current when the secondary connected is because inductance decreased and frequency is the same. From my example, the LC circuit would auto adjust its frequency to keep itself in resonance. The equivalent would be like variant two (decrease inductance but added inductance). But this means we cannot take energy out of the LC even when draw power from it... Hm...
Awesome pdf,
GH
Just saw your post MH. Thanks, but would this be the case?
(http://www.overunityresearch.com/index.php?action=dlattach;topic=1312.0;attach=7109;image)
Hi Groundloop,
The secret to this phenomenon you experienced is linked to two important aspects of your circuit configuration:
1. Lead Acid batteries are recharged by current not voltage (although voltage is a common component of precipitating current flow inside the battery)
2. Buck Boost operation is a cyclic process.
During the off period of the drive transistor the magnetic field in the transformer collapses and energy presented to the base drive winding (upper left in your drawing). Since it is not good for that energy to be presented to the base, it is routed back to the supply line via the upper most diode. This causes a reverse current (ever so small as it is) to flow your input battery.
Normally, the greater portion of that energy will be absorbed by the secondary and load - but when your output battery is fully charged, the input battery becomes the greater load and the energy is directed as stated above. There is also a secondary parasitic operation caused by current flowing in the secondary winding - this too builds a field that can be siphoned off through the path mentioned above. Since we are sloshing the energy in the secondary and blocking it on part of the cycle, it allows us to use energy from the secondary (output battery) to recharge the primary (input battery) on that part of the cycle.
Eventually the two batteries would find an equilibrium related to the turns ratios and impedances involved - less of course the losses from circuit operation.
As a side note regarding Poynt99's comment regarding professor Lewin - I disagree wholeheartedly. Professor Lewin did not make a mistake, and he clearly showed that Kirchhoff's Law is a subset and 'special case' of Faraday's law. Any engineer that understands path dependency will fully agree with these statements. Kirchhoff's law simply will not hold where path dependency is present and real world examples of this were previously provided in a prior thread with a link to a well written magazine article on that specific subject.
8)
Quote from: Magluvin on 2012.02.16, 23:49:33
Gibbs, read this pdf. Its not long, but it can answer your question as for one way of doing it. ;]
Mags
Mags,
Se my Figure-8 research on my bench. :-)
http://www.overunityresearch.com/index.php?topic=535.0 (http://www.overunityresearch.com/index.php?topic=535.0)
Regards,
GL.
Hi Harvey,
Nice to see you back on the forum again. :-)
Your explanation fits 100% with the observed circuit behavior.
Thanks,
GL.
Quote from: MileHigh on 2012.02.17, 00:09:53
WaveWatcher:
I don't know what's going on in your treatise. There is a diode restricting the direction of the current flow on the output battery so I don't see any mechanism for the energy to flow from the output battery to the input battery.
Harvey posted a much more precise explanation of the reverse charging function. I should not have used the term 'magnetic amplifier'. That tends to sound a bit esoteric to many.
And no. The diode is exactly as it should be for this to function.
Harvey,
Right on! on your post about the Lewin experiment. Kirchhoff is satisfied because his laws do not apply. His work was done when Faraday's laws over ruled him. You use the term 'sub-set' and that is completely correct. KVL only applies to path independent potential distribution. Path dependency is the telltale sign.
It is a very good thing this works as Lewin described or there would be dozens of different type devices that would never work.
Sorry .99. I would love to see your conclusive results some time. They must be very convincing.
I came up with almost exactly the same conclusion offered by Lewin a very long time ago. Since then, I've performed variations a couple of times. The conclusions never change.
Groundloop, Harvery, and I think WaveWatcher also,
It looks lo me like you guys are speculating on the fringe here. For starters, a way to really determine what's going on would be to simulate the circuit in PSpice. You just have to replace both batteries with 12-volt voltage sources, and perhaps add a little output impedance resistor to each voltage source to resemble batteries a bit more.
I stared at the circuit for 10 minutes and taxed my brain. Here are my thoughts.
Let's use the terminology primary battery (left) and secondary battery (on the right, the one that gets charged). We will say the secondary winding of the transformer is the 230-volt side, it charges the secondary battery.
For starters, the power stroke to charge the secondary battery is when the transistor switches on. All of the circuitry to the left of the transformer is an oscillator.
You can see how the dot on the 230-volt secondary is on the wrong side. In the schematic it's on the bottom when it should in fact be on the top. When the transistor switches on the "dot" on the lower primary winding goes low. So on the secondary winding the dot must be on the top, so that the other side of the secondary winding goes high and that forward biases the diode and charges the secondary battery.
And that's the essence of the whole design. The transistor switches on and while it is on you are in power stroke mode to charge the secondary battery. There is no magnetic energy in the core at the end of the power stroke at all. It all gets transferred into the secondary battery. It doesn't matter if the secondary battery is partially charged or fully charged, during the power stroke current flows through the secondary winding to charge the secondary battery.
Here is a big point that I think you are missing: During the power stroke, the lower primary winding is putting changing magnetic flux into the core. There are many turns on the secondary, the ratio from secondary winding to primary winding is about 30:1. Therefore, as fast as flux is being put into the transformer core by the primary, it's being "taken away" or cancelled by the secondary. The core is just an energy transfer medium in this case, it's not storing energy.
The top primary coil is just part of a feedback mechanism to keep the transistor on for a certain amount of time. It acts akin to what you see in a Joule Thief to snap the transistor on and keep it on.
When the transistor switches off, nothing special happens to the primary battery. There is never any energy from the secondary battery that makes it back to the primary battery.
If a very small amount of energy goes back and recharges the primary battery through the top diode, it's a moot point. Any energy that goes back to "recharge" the primary battery came from the primary battery in the first place.
Like I said before, the near-dead zombie battery on the primary is increasing in voltage because it is being "exercised," not because it is being recharged.
Anyway, that's my take on it. Standard disclaimers in that I could be wrong, but that's what I think I am seeing.
MileHigh
Just a few more comments about Groundloop's circuit as I have been taxing my brain some more. God, I am glad I never became an analog design engineer and I changed careers. You really have to "enjoy" this stuff and live it and breathe it to truly be competent in it. I am not there at all, not in the "zone" so I may not be correct. That's where PSpice can come to the rescue.
The key thing about this circuit is that the transformer is working in pulse mode. You apply a voltage pulse on the input and you get a voltage pulse on the output that goes into a load. The transformer is basically transferring a signal from the primary to the secondary. In other words, you can put a square wave voltage waveform on the input and you will see a square wave voltage waveform on the output.
In pulse/signal mode, if you have a 1:1 ideal transformer and a one-ohm load, then a one-volt positive pulse on the primary will become a one-volt positive pulse on the secondary. The input current when the pulse is active will be a constant one amp, and likewise the output current when the pulse is active will be a constant one amp. So you can see how the pulse-mode transformer can theoretically just pass a signal almost like it isn't there. The load on the secondary is reflected back to the primary perfectly (in an ideal case).
In this battery charging circuit the load on the secondary is non-linear, it's a battery. Also, we know that the secondary wants to create a very high EMF because of the very high turns ration of 30:1. But the secondary battery will clamp the output voltage from the secondary winding to 12 volts. That means that whenever the transistor pulses low and drives the transformer primary, the secondary is reflected back to the primary and looks like a "quasi short circuit." i.e.; the secondary winding wants to put out 360 volts but it is clamped to 12 volts by the secondary battery. This looks like a "quasi short circuit" as far as the secondary winding is concerned. This "quasi sort circuit" is reflected back to the primary. This "quasi short circuit" simply means that the pulse transformer under these conditions will suck as much power from the primary battery as is possible.
So when the transistor switches on and goes into power stroke mode, the secondary battery is supplied with as much power as possible. The primary battery sees a very low impedance load on the transformer primary, and it outputs a constant pulse of DC current with a flat top as long as the transistor is switched on. This puts a constant pulse of DC current through the secondary battery. Again, don't forget that this is a relatively short pulse.
When the pulse is on during the power stroke, there are three main agents that the power is flowing through. There is the primary battery, the transformer, and the secondary battery. One of these three agents will be the "power bottleneck" that regulates the maximum power flow rate during the power stroke.
Coming back to one of the main points, because the turns ratio in the transformer is so high, 30:1, when the transistor switches on it looks almost like a short circuit to the primary battery, and everything is "stressed" as you try to "over voltage" the secondary battery. It's impossible to over voltage the secondary battery and in reality it clamps the secondary coil EMF output to 12 volts. That clamping action is equivalent to a short circuit.
I don't know if this helped but that's what I am seeing. Because of the "quasi short circuit" effect, you would have to be careful with your PSpice simulation and add a few resistors where they are supposed to go (transformer windings, etc) in order to get an accurate simulation going of what is taking place in the real world circuit.
MileHigh
I'll chip in here just to say that IMO any charging effect of the primary battery is not as a result of a reverse current flow. I have seen this a few times and its due to battery condition where the battery appears to be charging when subjected to a load, especially when being pulse discharged. The duration of this effect is related to the condition of the battery and can last for a considerable time before the primary battery exhibits a normal discharge curve. Throughout the period that the primary battery is showing signs of charging, energy is in fact being taken from the battery to power the load. This effect has so often been misinterpreted as free energy in pulse motor / heater setups that are being powered from LA batteries. Having said this, there may be another reason why this effect can occur but the effect I describe is IMO the most likely explanation.
Regards
Hoppy
Quote from: Hoppy on 2012.02.17, 08:29:26
I'll chip in here just to say that IMO any charging effect of the primary battery is not as a result of a reverse current flow.
You changed your stance? Ok. Then we are in disagreement. The effect GL was asking about is actually known as a design flaw to be avoided. At least it is in my circles.
Yes, the severely weakened battery will appear to be charging when exercised. The voltage may actually increase but that doesn't mean the stored energy has increased. When this happens the battery impedance also increases. I agree that not understanding this may be the primary basis for the Bedini-ite existence.
GL,
If you run into the problem again a simple fix for this circuit is to add a current limiting resistor in the circuit of the third winding - the feedback/trigger winding for the transistor.
Spice is handy. I use some flavor of it frequently. The problem with it is the same as computer programming.
You don't write a computer program because you aren't capable of performing a task. You write it because you are capable but don't wish to perform the task or using a program is more cost effective, etc.
pSpice is much the same. It is best if you don't use it to find out what a circuit may do. The reason is you must correctly understand the circuit before you can build the virtual circuit.
pSpice will do whatever you program it to do, often even when what you believe is incorrect.
Quote from: WaveWatcher on 2012.02.17, 11:03:34
You changed your stance? Ok. Then we are in disagreement. The effect GL was asking about is actually known as a design flaw to be avoided. At least it is in my circles.
Yes, the severely weakened battery will appear to be charging when exercised. The voltage may actually increase but that doesn't mean the stored energy has increased. When this happens the battery impedance also increases. I agree that not understanding this may be the primary basis for the Bedini-ite existence.
Hi WW,
??? I did not suggest that the stored energy has increased. To the contrary, as I said it depletes over the duration of the 'effect' i described.
Regards
Hoppy
Hoppy,
Sorry. I took this...
Quote
There is also a secondary parasitic operation caused by current flowing in the secondary winding - this too builds a field that can be siphoned off through the path mentioned above. Since we are sloshing the energy in the secondary and blocking it on part of the cycle, it allows us to use energy from the secondary (output battery) to recharge the primary (input battery) on that part of the cycle.
.... as an alternate but valid explanation of the amplification performed by the transformer when acting as a magnetic amplifier during 1 half of the cycle.
There is a similar known problem in older signal amplifiers which used capacitive coupling. It was due to the electrostatic amplifier effect. Both, magnetic and electrostatic amplification are interests of mine and both are quite valid. It is possible for amplification or energy switching to go the wrong direction.
In poorly designed emergency battery chargers (using a rechargeable battery as the source) it is possible to drain the battery intended to receive charge and charge the source battery.
In poorly designed amplifiers (audio up to HF RF AFAIK) you can wind up with enough feedback to create an oscillator that can destroy intermediate stages. I haven't heard of this in several years so I must assume those highly electrostrictive capacitors are off the market.
QuoteIn poorly designed amplifiers (audio up to HF RF AFAIK) you can wind up with enough feedback to create an oscillator that can destroy intermediate stages. I haven't heard of this in several years so I must assume those highly electrostrictive capacitors are off the market.
This is exactly what is necessary...
Quote from: allcanadian on 2012.02.16, 20:31:53
@GibbsHelmholtz
That is an interesting question isn't it, I can take a pendulum on magnetic bearings in a vacuum and store energy as oscillations indefinitely, I can take a capacitor and charge it and store energy indefinitely however when I try to store energy as a static unchanging field in a coil of wire I must continually input energy to maintain this field despite Ampere's statement
...
There is a perfect correspondance between:
Coil
|
<-> Capacitor
|
Magnetic field
|
<-> Electric field
|
Current
|
<-> Voltage
|
Low series resistance
|
<-> High parallel resistance
|
Closed coil circuit
|
<-> Open capacitor circuit
|
The differential equations of circuits with capacitors or coils are the same, only the static parameters are different.
A perfect coil has a zero series resistance, and a perfect capacitor has an almost infinite parallel resistance.
So we understand that in practice we have almost perfect capacitors, because their parallel resistance is very high, especially for non electrolytic capacitors. Unfortunately, we have not the same chance with the coils, in which the series resistance is really not neglectible. If our every day coils were as perfect as our capacitors, they would be almost superconductor. If a circuit with such a superconducting coil was closed, you would need electrical energy only to build the magnetic field by generating a current in the coil and then, the current would flow indefinitely in the coil, in the same manner as a voltage is maintained indefinitely in a capacitor and you need energy only to build the electric field by charging the capacitor.
It is only the technology and not the principle, that makes the difference between coils and capacitors from the viewpoint of keeping the energy. Coils are very lossy components, capacitors are near ideal.
The (rather high) energy that you must continually input in the coil is that which is wasted in its (not low enough) resistance.
The (rather low) energy that you must continually input in a capacitor is that which is wasted in the (very high) leakage parallel resistance.
Quote from: exnihiloest on 2012.02.17, 16:44:56
There is a perfect correspondance between:
Coil
|
<-> Capacitor
|
Magnetic field
|
<-> Electric field
|
Current
|
<-> Voltage
|
Low series resistance
|
<-> High parallel resistance
|
Closed coil circuit
|
<-> Open capacitor circuit
|
The differential equations of circuits with capacitors or coils are the same, only the static parameters are different.
A perfect coil has a zero series resistance, and a perfect capacitor has an almost infinite parallel resistance.
So we understand that in practice we have almost perfect capacitors, because their parallel resistance is very high, especially for non electrolytic capacitors. Unfortunately, we have not the same chance with the coils, in which the series resistance is really not neglectible. If our every day coils were as perfect as our capacitors, they would be almost superconductor. If a circuit with such a superconducting coil was closed, you would need electrical energy only to build the magnetic field by generating a current in the coil and then, the current would flow indefinitely in the coil, in the same manner as a voltage is maintained indefinitely in a capacitor and you need energy only to build the electric field by charging the capacitor.
It is only the technology and not the principle, that makes the difference between coils and capacitors from the viewpoint of keeping the energy. Coils are very lossy components, capacitors are near ideal.
The (rather high) energy that you must continually input in the coil is that which is wasted in its (not low enough) resistance.
The (rather low) energy that you must continually input in a capacitor is that which is wasted in the (very high) leakage parallel resistance.
Thanks for the explanation. I didn't aware of the parallel resistance in cap. Is this the same as capacitor internal resistance?
Quote from: Groundloop on 2012.02.17, 02:07:55
Mags,
Se my Figure-8 research on my bench. :-)
http://www.overunityresearch.com/index.php?topic=535.0 (http://www.overunityresearch.com/index.php?topic=535.0)
Regards,
GL.
Nice. :]
I see your using it in a different way. Ill look at it tonight to see what your up to. ;]
What I get from the pdf is when you set up the primary for resonant oscillation, how ever one might do, and the secondary with the additional core wont spoil(kill) the primary oscillation. So now, we can draw energy from the oscillation(pendulum) while easily maintaining the primary oscillation(pendulum) with the input.
Mags
Quote from: GibbsHelmholtz on 2012.02.17, 17:23:15
Thanks for the explanation. I didn't aware of the parallel resistance in cap. Is this the same as capacitor internal resistance?
I presume it is the same. I called it "parallel resistance" Rp because in the schematic of a real capacitor, it is put in parallel with the cap C. There is also a resistance Rs in series (the resistance of the wire and of the electrodes) but this one is generally not disturbing except for high currents, and of course it plays no role when the capacitor circuit is open. And there is even a "coil" in a capacitor: the inductance L due to the wire and the electrodes. In despite it is low, it is not negligible for high frequencies.
So a real capacitor is: Rs+L+C//Rp. The lower Rs, the lower L, the higher Rp, the better the capacitor.
A real coil is : Rs+L//C//Rp. Rs is the resistance of the wire, L the inductance, C a parallel capacity due to each wire turn capacitively coupled to the next and the previous (LC produces the "natural resonance of the coil" at some frequency) and Rp is generally negligible (=infinite: resistance between turns due to bad insulation).
The lower Rs, the lower C, the higher Rp, the better the coil.
For the losses, Rs in the coil plays the role of Rp in the capacitor (and vice versa).
A little digression: when you have several coils, you must take into account couplings not only inductive but also capacitive due to the capacity between coils, each one acting as a capacitor electrode. The number of parameters and of resonant modes become high, and this explains the miscellaneous things happening in a TPU. There is no mystery, but only a high number of real parameters which are difficult to deal with, because they have no obvious visibility and we don't know the values. In the real life, it is not as in the free energy life, capacitors are not capacitors and coils are not coils :). These are the every day problems of the engineers.
Quote from: GibbsHelmholtz on 2012.02.17, 17:23:15
Thanks for the explanation. I didn't aware of the parallel resistance in cap. Is this the same as capacitor internal resistance?
The parallel resistance is quite high and is
the leakage resistance of the dielectric
within the capacitor.
The series resistance (ESR) is quite low and is
the resistance of the metallic plates and leads
(conductors) of the capacitor.
Thanks guys :)
Quote from: MileHigh on 2012.02.16, 23:48:18
Farmhand:
Your questions can't be answered. As a bare minimum you need a schematic and a description of what you think your circuit is doing. Preferably you would include scope shots.
Ok, I will do that as soon as I can, I need to make a schematic anyway, I'll have to video tape the scope and take capture's as I don't have a scope that stores data, thanks for the reply.
Cheers
Okay so let's revisit Min2oly's clip and I will give you my explanation of what's going on. Like I said several people have given reasonably good explanations but there was no "Bingo!" moment as far as I am concerned. Nobody attempted to explain why the capacitor reaches a given voltage either.
When he first makes the connection to the coil with the alligator clip that's at ground potential, no current flows at the instant of connection. Current starts to flow as the coil gets energized. Some battery power is also being burned off in the resistance of the coil. When he removes the alligator clip the amount of energy stored in the coil is 1/2Li-squared.
The instant the alligator clip is removed then the current through the coil does not change at all. Instead it flows through the diode and charges the capacitor. Now the coil is starting to discharge its stored energy. It's important to restate that when the coil is discharging (a.k.a. magnetic field collapsing) it's a current soure, the operating principle is the flow of current, not the generation of voltage.
The current flowing out of the coil flows through the diode and starts to charge the capacitor. It's the capacitor that generates the voltage, V=Q/C. The coil then reacts to the fact that the capacitor has a certain voltage across it and will generate sufficient EMF to overcome the capacitor voltage plus the 0.6 volt voltage drop of the diode to keep the current flowing. This is an important distinction that should be appreciated.
Eventually the coil runs out of stored energy and the current stops flowing. So the voltage in the capacitor is the 6 volts from the battery plus whatever contribution came from the discharging coil. That can be easily calculated. We can ignore the energy lost in the diode and the wire resistance to keep things simple for illustrative purposes. The energy stored in the coil is 1/2Li-squared. This same amount of energy is transferred into the capacitor, where the formula is 1/2Cv-squared. So if you know the inductance of the coil and the initial current flow, and you know the capacitance, then you can calculate the voltage that will end up across the capacitor. i.e.; 1/2Li-squared = 1/2Cv-squared. You have that equation so you can solve for v, the voltage that ends up in the capacitor when Min2oly breaks the connection to ground by removing the alligator clip.
The moral of the story is two-fold: It's clear that Min2oly is totally wrong in this "teaching" clip, and that probably comes from the Yahoo Bedini groups. You have hundreds of people that are misled. The other thing is the only reason you are getting high voltage across the capacitor is because the capacitor is very small. If you used a much larger capacitor in the clip, the "radiant energy spike" would completely disappear.
There is no "radiant energy" being demonstrated here at all. What is being demonstrated is battery energy first going into the coil, and then being transferred into the capacitor. The resistance of the multimeter then bleeds off the charge in the capacitor and we see the voltage decrease.
MileHigh
There are a few follow-up pop quiz tech questions that came earlier in the thread and I will add a few others myself.
1. We know that if you replace the charging battery in a Bedini motor setup with a large capacitor and a resistor, that you can make a trivial measurement to measure the average output power from a Bedini motor's discharging drive coil. P_average_out = v-squared/R. The value of the resistor doesn't matter, within limits. Note that you never see this done by your typical Bedini motor builder so they usually don't have an accurate measurement of the average output power from the coil. It's a fundamental measurement that should not be ignored. Some people actually believe that it's impossible to make this measurement.
All that being said, there are two questions:
1.1 If you want to make as accurate a measurement as possible of the output power, you should adjust the value of the resistor so that the voltage across the capacitor stabilizes at about 12.6 volts. Why is this?
1.2 If the voltage across the capacitor is very low, your measurement can be invalid. Why is this?
2. With the average output power setup discussed in point 1 above, how can you use this information to make a quite accurate measurement of the inductance of the main drive coil in your Bedini motor?
3. Going back to my explanation in the previous posting, calculating the true final voltage in the capacitor (still ignoring the energy loss in the coil resistance and diode energy loss) is still a bit complicated because of the presence of the 6-volt battery in the circuit. What issues might come into play?
4. When the alligator clip is disconnected I stated that the current flowing through the coil remains constant and now starts to flow through the diode and into the capacitor. In reality the current flowing through the coil actually decreases by a very very small amount. It's takes an instantaneous "step" down to a slightly lower level. Why is that?
Now if any pulse motor builders give me attitude about "not being a builder" I can point them to this thread!
MileHigh
Quote from: MileHigh on 2012.02.18, 23:08:16
1.1 If you want to make as accurate a measurement as possible of the output power, you should adjust the value of the resistor so that the voltage across the capacitor stabilizes at about 12.6 volts. Why is this?
Because this is inline with the recharge battery voltage?
Quote
1.2 If the voltage across the capacitor is very low, your measurement can be invalid. Why is this?
Because the current reading could be higher, which is inaccurate?
Quote
2. With the average output power setup discussed in point 1 above, how can you use this information to make a quite accurate measurement of the inductance of the main drive coil in your Bedini motor?
I can only see that use the power I^2R = 1/2LI^2 to solve for inductance. Not correct unit but because average power is all the info we have.
Quote
3. Going back to my explanation in the previous posting, calculating the true final voltage in the capacitor (still ignoring the energy loss in the coil resistance and diode energy loss) is still a bit complicated because of the presence of the 6-volt battery in the circuit. What issues might come into play?
Battery voltage assist the discharge giving it more energy than what should be?
Quote
4. When the alligator clip is disconnected I stated that the current flowing through the coil remains constant and now starts to flow through the diode and into the capacitor. In reality the current flowing through the coil actually decreases by a very very small amount. It's takes an instantaneous "step" down to a slightly lower level. Why is that?
Because of the diode voltage drop .6V which oppose the ideal 0V initial current ?
I need partial credit, this is how I did not fail classes. lol
Gibbs:
For your answer to 1.1, I forgot to mention you set the capacitor voltage to 12.6 volts to be the same as the charging battery voltage. That is a given. So although what you say is correct, I am looking for another answer.
For 1.2, you are in the right area, but one more time, you are not giving the proper answer.
For 2, 3 and 4 no partial credit, sorry!
I just want to revisit the point about the capacitor actually being the agent that is responsible for the voltage generation, it's not actually the coil itself. Another idea worth repeating is that the coil will simply force it's stored energy into the capacitor no matter how high the voltage has to get. If you pretend that you have an ideal (perfect) diode between the coil and the capacitor, then in theory a very very small capacitor could get charged to 20,000 volts or higher. Again, the coil is "uncaring" about the voltage, it just wants to try to maintain the current flow. In reality, the current flow drops to zero as the coil discharges. While this is happening, the voltage output by the coil could be anything. The voltage generated by the coil is dependent on the nature of the load that the coil is discharging into. That explains the so-called "radiant spikes" which are not radiant spikes at all.
If you are stuck and you can't understand this concept devote as much energy and time as you can to getting over the hump and understanding it. You can't build Bedini and other pulse motors and really know what you are doing unless you can understand this concept. Think of the poor yahoos on the Yahoo Bedini forums. Between not understanding how coils work and fantasizing about "radiant energy" and measuring battery voltages and falsely believing that this is an indication of the battery state of charge, they are like hopelessly lost sheep.
Also, I did not mention the core vs. no core. Most know that when you add a core you increase the inductance of the coil. So the coil can store more energy for a given current flow. Hence more energy can be transferred into the capacitor and that's why you get higher voltages.
I always like to think of a coil with a core as a kind of "two layer cake." For zero current to core saturation current, you have a large inductance coil. Above the saturation current level, you are back to a small inductance coil, the "air coil." The starting point for calculating the energy in the "air coil" is at "zero" at the saturation current level. So it's like a two-layer cake and the total inductance for current above the saturation current level is both layers added together.
MileHigh
One last comment about the coil and the fact that it is outputting current and not "responsible" for the generation of voltage. I am stating that as a way to hopefully make it easier to understand what is going on.
Somebody might come along and beat me up about that statement because it's not literally true.
On a more technical level, both capacitors and coils generate a combination of voltage AND current when they discharge their stored energy. They have to by definition, energy is voltage times current times time.
However, for coils it's fair to state that the nature of the load determines the voltage output by the coil. For capacitors it's fair to state that the nature of the load determines the current output by the capacitor.
The hard-core equation is: v = L di/dt.
Or the same equation with more detail: v(t) = L di(t)/dt
"The voltage generated by an inductor is the product of the inductance of the inductor and the rate of change of current flowing through the inductor with respect to time."
It's this part, "the rate of change of current flowing through the inductor with respect to time" that is determined by the load on the coil when it is discharging.
i.e.;
high load resistance = faster energy burn = high rate of change of current flowing through the inductor with respect to time = high output voltage.
low load resistance = slower energy burn = low rate of change of current flowing through the inductor with respect to time = low output voltage.
zero load resistance = no energy burn = no change in the current flowing through the inductor with respect to time = zero output voltage.
A very small capacitor sort of looks like a very high load resistance, hence it charges to a very high voltage.
If you get that then you really know what you are talking about. Figure out the same principles for capacitors and then you are starting to get "real" when it comes to electronics.
MileHigh
Quote from: MileHigh on 2012.02.19, 06:09:19
Gibbs:
For your answer to 1.1, I forgot to mention you set the capacitor voltage to 12.6 volts to be the same as the charging battery voltage. That is a given. So although what you say is correct, I am looking for another answer.
For 1.2, you are in the right area, but one more time, you are not giving the proper answer.
For 2, 3 and 4 no partial credit, sorry!
MileHigh
Damn! This is the kind of result I get. Actually, this one is not too bad. lol
MH,
When I read about people doing Bedini motor setup, they said they can recover 80- 98%. Some even claims that they have >100%. Is this even possible? What is the highest they can go if we ignore all losses? Thanks
Gibbs:
QuoteWhen I read about people doing Bedini motor setup, they said they can recover 80- 98%. Some even claims that they have >100%. Is this even possible? What is the highest they can go if we ignore all losses?
Sorry, this is not a sensible question.
The real question in this: If I put energy into my Bedini motor with a source battery and charge up a charging battery, how much of the source battery energy can I get back from the charging battery?
The answer is probably about 20%. This might seem shocking but the problem is that nobody does their battery measurements properly. They have to measure the
energy in their batteries and they never do it.
Here is the explanation, and we will use 10,000 Joules of energy for illustrative purposes.
1. Source battery puts 10,000 Joules of energy into the Bedini motor.
2. 7,000 Joules of energy lost as heat, 3000 Joules of energy go into the charging battery.
3. Assume charging efficiency of charging battery is 0.8 -> 2400 Joules become chemical energy inside battery and 600 Joules lost as heat.
3. Assume discharging efficiency of charging battery is 0.8 -> 1920 Joules of energy available to drive an electrical load and 480 Joules lost as heat.
Efficiency is 1920/10,000 x 100 =
19.2% efficiency.These are perfectly reasonable numbers. A typical Bedini enthusiast would be shocked to hear this and might find it very offensive. The problem is that they have never made the proper measurements. I have observed enough Bedini clips to know with a high deal of confidence that a typical Bedini motor is only 30% efficient. I am not a battery expert, but I believe that the charging and discharging efficiencies that I am quoting for a lead-acid battery are reasonable.
All the talk about the "radiant spike" getting the battery to "resonate" and "get extra energy from the environment" is simply junk.
Any Bedini motor enthusiast can make the measurement to determine the efficiency of the motor in step 2 above, it's clearly explained in this thread. Then once they have that they have to make proper battery energy measurements to determine the combined (charging + discharging) efficiency of the charging battery. I am estimating the combined efficiency for a typical lead-acid battery would be about ( 0.8 x 0.8 ) = 0.64. Unfortunately, it's highly unlikely these measurements will ever be done.
MileHigh
Quote from: MileHigh on 2012.02.20, 03:11:28
Gibbs:
Sorry, this is not a sensible question.
The real question in this: If I put energy into my Bedini motor with a source battery and charge up a charging battery, how much of the source battery energy can I get back from the charging battery?
The answer is probably about 20%. This might seem shocking but the problem is that nobody does their battery measurements properly. They have to measure the energy in their batteries and they never do it.
Here is the explanation, and we will use 10,000 Joules of energy for illustrative purposes.
1. Source battery puts 10,000 Joules of energy into the Bedini motor.
2. 7,000 Joules of energy lost as heat, 3000 Joules of energy go into the charging battery.
3. Assume charging efficiency of charging battery is 0.8 -> 2400 Joules become chemical energy inside battery and 600 Joules lost as heat.
3. Assume discharging efficiency of charging battery is 0.8 -> 1920 Joules of energy available to drive an electrical load and 480 Joules lost as heat.
Efficiency is 1920/10,000 x 100 = 19.2% efficiency.
These are perfectly reasonable numbers. A typical Bedini enthusiast would be shocked to hear this and might find it very offensive. The problem is that they have never made the proper measurements. I have observed enough Bedini clips to know with a high deal of confidence that a typical Bedini motor is only 30% efficient. I am not a battery expert, but I believe that the charging and discharging efficiencies that I am quoting for a lead-acid battery are reasonable.
All the talk about the "radiant spike" getting the battery to "resonate" and "get extra energy from the environment" is simply junk.
Any Bedini motor enthusiast can make the measurement to determine the efficiency of the motor in step 2 above, it's clearly explained in this thread. Then once they have that they have to make proper battery energy measurements to determine the combined (charging + discharging) efficiency of the charging battery. I am estimating the combined efficiency for a typical lead-acid battery would be about ( 0.8 x 0.8 ) = 0.64. Unfortunately, it's highly unlikely these measurements will ever be done.
MileHigh
That's what I thought too MH. So somewhere in the inductor transient explanation you gave, the wheel magnet coming in and take energy. I've never heard the motor group explain how it effect the recovery process.
Interesting how Min2oly is now trying to equate a bloch wall and a coil. Note the title of his new clip, "Bloch Wall compression and extraction of free energy."
Of course there no bloch wall associated with a coil or for a standard bar magnet. In the above clip Min2oly is basically doing the same thing that he did in his previous clip. There is one somewhat helpful comment on his YouTube clip and the rest are not helping him at all.
About 1 1/2 years ago I made a single technical comment on one of Min2oly's clips and he instantly banned me.
If somebody could do me the favour of pointing Min2oly to this thread that would be appreciated. I can't, my hands are tied.
MileHigh
So what. Are you your brother's keeper?
He uses a dvm as a counter and believes it. Even though the meter is not designed for the fast connections.
Don't correct him. Let him babble. I cant even listen to the daytime soap opera music in the background. He has been watching too many Bela Lugosi movies. >:-)
The vocabulary and the technical terminology
which Min2Oly speaks in his video reveals
much.
His "education" is just beginning and at this
stage most of what he's "learned" is from
his internet peers.
Hopefully he'll one day soon delve into the
"books" and do some real research and study.
Then he'll learn the correct terminology and
be able to speak with credible knowledge as
he explains what is really taking place in his
demonstration.
The word of mouth (or word of forum) errors
which are handed down from one person to
the next while blindly being accepted as "fact" are
truly astonishing.
The Art of Study is seemingly becoming a lost
skill as so few of the forum experimenters actually
conduct any meaningful research on their own.
It is yet unknown as to why this is; they are either
incapable, insufficiently motivated, or lack the ability
to read and comprehend well.
Perhaps it's a result of the "dumbing down"
which education American Style has become.
But, then again, it's not just in America is it?
The "Bedini" threads at Energetic Forum are a
convincing example of the blind leading the blind...
http://www.youtube.com/watch?v=fMkrJOOTjj4
Giantkiller:
What's wrong with trying to help someone learn something? I realize that it's not as simple as that, but still the fundamental point about wanting him to learn and understand is true. It's might upset the apple cart and cause some consternation in certain circles. The is a sweet rebut to that though. The rebut is that's what certain circles think they are doing themselves and they are proud of it. So if that upsets the apple cart of the upsetters that want to upset the apple cart, you have kind of gone full circle. I don't think the two Universes will annihilate each other if a connection is made between the two like in that old Star Trek episode.
But, it will take some courage on the part of somebody to tell him and point him to this thread. I know from direct experience that the peer group pressure in the free energy realm among forty-something men is just as strong as it is among a bunch of overly intense girls in grade nine high school.
To quote Min2oly:
Quoteso this collapsing inductive spike is not CEMF either.
where does the spike come from?
what is the conventional law that explains it?
I'm being very sincere with my question.
Thanks,
Patrick
The answers to his questions are in this thread.
Dumped: Certainly we share some common ground with respect to this subject matter.
I would be nice to see the truth make some inroads. People on this forum that straddle "both sides of the divide" that know Patrick can decide if they want to do it or not.
MileHigh
What the hey.... Here is a big Easter Egg for Min2oly if he actually reads through this thread from start to end and gets it...
Call this the "Holy Grail" for all of the Bedini motor enthusiasts. Oh my Gawd!!!! All three cases explained for the amazing (voltage ???) spike that has fascinated generations.....
Finally, we reveal all..... Time to expose the wizard and get back to Kansas.
So how does this discharging inductor explanation apply to a Bedini motor?
>>> Case 1: The charging battery is connected.
When the transistor switches on, the current flowing through the coil starts to increase.
When the transistor switches off, the current that was formerly flowing through the coil now continues flowing through the diode and into the charging battery. If the charging battery is in good condition but in a discharged state, there is just a mild bump up in the battery voltage. There is only a mild bump up in the battery voltage due to internal charging resistance of the battery. We know that the internal charging resistance of a healthy battery is quite low, hence a mild bump up in the voltage.
>>> Case 2: The charging battery is disconnected.
When the transistor switches on, the current flowing through the coil starts to increase.
When the transistor switches off, the current that was formerly flowing through the coil now continues flowing through the neon. The neon reacts to the incoming current flow by ionizing it's gas and doing a "controlled burn" where the voltage is at about 90 volts. This represents a fairly high power dissipation rate. For example if the current flowing through the coil is one ampere when the transistor switches off, and the breakdown voltage of the neon bulb is 90 volts, then the initial "burn power" is 90 watts. The energy stored in the coil is extinguished fairly quickly.
>>> Case 3: The charging battery is disconnected and there is no neon to protect the transistor.
When the transistor switches on, the current flowing through the coil starts to increase.
When the transistor switches off, the discharging coil literally forces current through the transistor as it is in the process of switching off. As the transistor is in the process of switching off, the effective collector-to-emitter resistance is constantly increasing. That's not stopping the coil by any means, and the more the transistor switches off, the higher the coil raises its voltage to maintain the current flow. This results in a very short and nasty complete discharge of the coil's stored energy inside the transistor itself. If the transistor switches off fast enough the coil will rip the innards of the transistor to shreds.
MileHigh
Well the thread has been idle for 120 day;s,but as i am a new member and have only just read it-i would like to look at this topic again.
I must say ,i find nothing wrong with or how min2oly explained the event's he was seeing or showing.
I also think calling it the inductive kickback is wrong-if anything ,it should be called an inductive kick forward-as the current dosnt change direction.
So why is min2oly's description of calling it the radiant spike wrong?
And befor you get to excited-im in no way a ###### fan-infact,up the garden path come's to mind when that name is mentioned.
So,text book explination is -inductive kickback-cool
But what is radiant energy?
I think when you know the answer to this,you may find that min2oly was very close to being correct when he stated it was the radiant spike that is charging the cap.
20 to 30% back to the charge battery (if useing an SLA)may be the case,but with a resistive load on the output-70% return is no problem.
Oh and that dosnt take into account the mechanical rotation either-which most people tend to ignore.
Hey guy's you all know me, yea its that agitator :D
The emoticons over here a great ;D lol
Anyway thought I would take a stab at the topic C.C
http://makeagif.com/i/oud1Sa