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Author Topic: Bedini 10-Coil Alternative Discussion  (Read 66447 times)
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Poynt:

Thank you for adding that clarification.  It's easy to forget and leave out all of the details, so your added commentary is appreciated.

Fausto:

To add to Poynt's comments, there is a rule about discharging inductors.  The rule is that when they start to discharge their energy, the initial current flow is the same value as the instant before the discharge starts.

For example, suppose an inductor has one ampere of current flowing through it, and the low impedance is 10 ohms and the high impedance is 100 ohms.  With those initial conditions, let's look at Point's comments were we "plug in the values:"

load impedance 100 ohms = pulse voltage: 100 volts, voltage/current pulse duration: short, current pulse amplitude: 1 ampere
load impedance 10 ohms = pulse voltage: 10 volts, voltage/current pulse duration: long, current pulse amplitude: 1 ampere

In both cases above the pulse shape is like an exponential decay.  You probably have seen this shape for the voltage decay when you discharge a capacitor on your scope.  It's the same shape for a discharging inductor.  Note also that in both cases exactly the same amount of energy is being discharged from the coil.

It's worth mentioning that the rule for a discharging capacitor is that when it starts to discharge, initial voltage level is the same value as the instant before the the discharge starts.

MileHigh
   
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Fausto:

Let's discuss the power issues and impedance matching issues for the two examples in the previous posting; a 10-ohm load and a 100-ohm load.

Let's assume that for both cases the pulse frequency is one pulse per second.

In both cases the average output power is the same.

We assume that for both the 100-ohm load and the 10-ohm load the inductor discharges all of its stored energy before the next pulse arrives.  Therefore by definition the average output power has to be the same.

This confirms the statement that I have said many times in this thread:

The average output power output for a periodically discharging inductor is independent of the load.

Here is another way of stating the above property for a periodically discharging inductor:

Impedance matching does not apply for a periodically discharging inductor, like in a Bedini motor.

In a Bedini motor the condition of the charging battery will not affect the motor's ability to pump power into the battery.  The battery can be an old and highly sulfated battery with a high input impedance or a brand new battery with a very low input impedance.  It does not matter, the Bedini motor will be able to pump the same average power into the battery independent of the input impedance of the battery.  The same thing applies to the interconnect wires.  The thickness of the wires will affect the ohmic losses in the wire itself, but besides that there are no special impedance matching requirements associated with the wires because there are no impedance matching issues associated with the output of a Bedini motor.

Earlier in the thread I explain how the output impedance for a discharging inductor is actually infinity.  That's a difficult concept to understand.  For more information just go through the thread and do a string search on "infinity."

We know that for the 100-ohm load the pulse is shorter in duration.  Therefore, for the higher impedance 100-ohm load the average power of the pulse only is high.  For the lower impedance 10-ohm load the pulse is longer in duration so the average power of the pulse only is low.

Finally, you ask yourself the question, what happens to the average power of the pulse only as the impedance gets higher and higher?   The answer is that the the average power in the pulse only gets higher and higher.

What does this look like in real life?  I am sure that you have seen on your bench or you have seen YouTube clips where someone open-circuits a coil that has current flowing through it.  You see a spark go through the air.  That is the manifestation of the load impedance getting so high that the coil generates a very very short discharge pulse at a very very high voltage for a very very high average power for the pulse only.  In other words, all of the energy stored in the coil is discharged in a split second at an extremely high average power level.

Note one more time that there is a equivalent for this situation with a charged capacitor.  If you make the load on the capacitor zero ohms then it discharges all of its stored energy with a very very high pulse of current over a very very short amount of time at an extremely high average power level.

MileHigh
   
Group: Guest
I stumbled across this John Bedini posting so I will make some comments:

>>>
--- In Bedini_SGatyahoogroups.com, "john_bedini" <john_bediniaty...> wrote:
>
> Answer to some questions,
> The magnets around the wheel are only used for a trigger signal. What
> is driving the wheel is hidden from your view, what is hidden from
> your view is also the charging signal. The driving force of the wheel
> is scalar or magnetic south poles between the north poles. Make
> yourself a timing light by taking a green or red led with a 330 ohm
> resistor in series with it. Place skinny white strips down the center
> of the magnets around the wheel, connect the led across the coil and
> then tell me where the coil pulse is and what is driving the wheel.
> The force that is driving the wheel is the same force charging the
> battery. Do the test take one fully charged battery and one
> discharged battery, hook them up and see if you get one to one if you
> do you have just seen a unity machine, but please do not leave out
> the wheel rotation in you calculation, mechanical power is equal to
> work done, its a figure of 29% so what kind of machine have you built?
> You will find that the scalar south is driving the wheel and not the
> north pole.
> John
>>>

Quote
> The magnets around the wheel are only used for a trigger signal. What
> is driving the wheel is hidden from your view, what is hidden from
> your view is also the charging signal. The driving force of the wheel
> is scalar or magnetic south poles between the north poles.

The magnets around the rotor wheel actually serve two functions:  1)  to generate the trigger signal,  2)  to act as one half (passive half) of the drive mechanism that makes the rotor turn.  The other half is the energized drive coil which acts like an electromagnet and forms the second half (active half) of the drive mechanism.

Quote
The driving force of the wheel is scalar or magnetic south poles between the north poles.

The "scalar" or "virtual south" pole on the rotor does not in any way whatsoever act as the driving force of the wheel.  The "lack of a north pole" "looks like" a south pole to the pick up coil or for an external generator coil.  It's because electro-motive force is generated in the pick-up coil by changing magnetic flux with respect to time.  When a rotor magnet is moving away from the pick-up coil that results in the pick-up coil "seeing" decreasing magnetic flux with respect to time.  That "looks like" an approaching south pole.  The rotor has to be moving for these effects to be seen.  The word "scalar" is meaningless in this sentence.  

Quote
> The force that is driving the wheel is the same force charging the battery.

This is correct.  To be more detailed the force that is driving the wheel is the energy being provided by the source battery that is turning the drive coil into an electromagnet.  At the same time some of the source battery's energy is being stored in the inductance of the coil.  When the transistor switches off that stored electrical energy is discharged through the diode into the charging battery.  Make no mistake about it, the energy in the spike comes from the source battery, and from nowhere else.

Quote
Do the test take one fully charged battery and one
> discharged battery, hook them up and see if you get one to one if you
> do you have just seen a unity machine, but please do not leave out
> the wheel rotation in you calculation, mechanical power is equal to
> work done, its a figure of 29% so what kind of machine have you built?

You will not get one to one.  We know that the charging battery current is typically about 25%-35% of the source battery current consumption for approximately the same source and charging battery voltages.  That is telling you right there that the Bedini motor is not efficient at all in charging the charge battery and is an underunity machine.  If you work out a real before/after battery energy measurement technique you will be able to prove this for yourself.

Bedini's comments about the 29% mechanical power out are pure nonsense.  The mechanical power output from a Bedini motor is zero.  The spinning rotor is equivalent to a resistor pouring energy down the drain and turning it into heat.  In the real wold you do NOT call that mechanical output power.  You absolutely must put a real-world external mechanical load on the Bedini motor's rotating shaft to say that you are getting mechanical power out from the Bedini motor.  Do not let anybody including John Bedini pull the wool over your eyes with respect to this very important issue.  This is explained earlier in this thread in more detail.

Quote
> You will find that the scalar south is driving the wheel and not the north pole.

The "scalar" south is doing no such thing.  That's another nonsensical statement.  To repeat, what is driving the motor is the drive coil acting as an electromagnet interacting with the magnets on the rotor.  That is the only thing that is driving the rotor.

MileHigh
   
Group: Guest


Bedini's comments about the 29% mechanical power out are pure nonsense.  The mechanical power output from a Bedini motor is zero.  The spinning rotor is equivalent to a resistor pouring energy down the drain and turning it into heat.  In the real wold you do NOT call that mechanical output power.  You absolutely must put a real-world external mechanical load on the Bedini motor's rotating shaft to say that you are getting mechanical power out from the Bedini motor.  Do not let anybody including John Bedini pull the wool over your eyes with respect to this very important issue.  This is explained earlier in this thread in more detail.

The "scalar" south is doing no such thing.  That's another nonsensical statement.  To repeat, what is driving the motor is the drive coil acting as an electromagnet interacting with the magnets on the rotor.  That is the only thing that is driving the rotor.

MileHigh

So all fans designed to push air have zero mechanical power and do no work besides dissipating heat like a resistor? Seriously Milehigh, can you not see that mechanical work is done? Ive seen a ride on lawn mower DRIVEN by one of these machines, not at any great speed, but it is actually moving with a person on it. I dont care if it only goes 1 kilometer an hour to me that is mechanical power. It takes mechanical power to move 250 odd kilos of weight along a road. At the very least you should be able to see that placing a coil next to the rotor will allow you to generate SOME current in a conventional generator fashion. Surely this power cannot be there without some form of mechanical power.

I thought this would have been obvious....even for you.

Regards
   
Group: Guest
Ren:

You are jumping the gun again.  Watch, I am quoting myself and then I will quote you:

Quote
You absolutely must put a real-world external mechanical load on the Bedini motor's rotating shaft to say that you are getting mechanical power out from the Bedini motor.

Quote
So all fans designed to push air have zero mechanical power and do no work besides dissipating heat like a resistor?

A fan is a mechanical load on the rotor!  So we are agreeing with each other!  The fan puts a mechanical load on the rotor shaft - torque times angular velocity equals mechanical power.  That is useful work from the motor.

The issue is when you stare at a spinning Bedini motor all by itself, just spinning, and someone like Bedini says something like, "The mechanical puts it into over unity territory if you add it to the charging."  The implication being that just staring at a spinning Bedini motor it is doing mechanical work.  That's not the case.  Adding a fan changes the situation, and then the motor is doing real mechanical work.

Quote
At the very least you should be able to see that placing a coil next to the rotor will allow you to generate SOME current in a conventional generator fashion.

Of course I agree with that.  In fact on the Bedini measurement thread I got into a discussion on how to properly measure the output from a generator coil setup.

MileHigh
   
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Ren:

You are jumping the gun again. 
MileHigh

Sorry, but I dont think I am at all.

I saw this quote straight from your message:


Bedini's comments about the 29% mechanical power out are pure nonsense.  The mechanical power output from a Bedini motor is zero.

MileHigh

In my world when you say the mechanical power output is zero, that is what you mean. You seem to contradict yourself in the very next message.

Regards

   
Group: Guest
Just for fun, some comments about where the electrical energy goes that you pump into a Bedini motor.  These are very generic comments that apply to any motor.

The sound that you hear when the motor is turning could in theory be measured and you could say that the motor is putting out a certain average amount amount of sound power in watts.  The sound bounces off the walls and floor and ceiling and other objects in the room.  The sound obviously disappears so what happened to the energy associated with that sound power?  The answer is that the walls and other surfaces are "lossy" and as a result the sound heats up the floor, ceiling and walls.  So all of the sound energy becomes heat energy.

Heat energy is kind of like the "end of the line" and is the last stop as a place for the energy to go.

The motor makes your table or bench vibrate with a certain amount of vibrational power in watts.  It's the same deal for this vibrational power.  It is eventually dissipated and becomes heat power.  So it's accurate to say that while your Bedini motor is running it is heating up your workbench.

All of the friction power associated with the bearings becomes heat power and that heats up the bearings, which then heats up the air.

All of the air friction due to the spinning rotor also becomes heat power and that heats up the air.   This one might not be too obvious to some of the readers but it's true.  If you wave your arms in the air, you actually end up heating up the air.

This brings up back to what happens when you connect a fan to the shaft of a Bedini motor.  You are indeed getting a certain amount of mechanical power in measurable watts out of the Bedini motor in terms of torque times angular velocity.  That mechanical power spins a fan that pushes a certain amount of air per second at a certain velocity.  That looks just like air friction or the waving of your arms.  So that useful work output by the Bedini motor ultimately becomes heat power and heats up the air.

We also of course know that all of the electrical power that is dissipated in the resistance of the coils, wires, transistor, diode, etc, becomes heat power also, and that heats up the air.

So you look at all of the above and you notice that everything becomes heat power in the end.  That's why when you look at a Bedini motor and if you factor out the power that is being pumped into the charging battery, a spinning Bedini motor with no mechanical load on it is a glorified resistor.  Except for charging the battery, there is no mechanical output.  Therefore a standard Bedini setup does two things with the energy supplied by the source battery, 1) it produces heat, and 2) it charges the charging battery.

Note that I am intentionally listing the production of heat power first, and charging battery power second.  That's because more of the power supplied by the source battery is becoming waste heat as compared to the power that is going into the charging battery.

MileHigh
   
Group: Guest
Ren:

Quote
In my world when you say the mechanical power output is zero, that is what you mean. You seem to contradict yourself in the very next message.

I am not contradicting myself.  I am scrutinizing what Bedini is saying and I have seen it repeated by many Bedini experimenters.  This observation is made by Bedini and other experimenters without ever saying that some sort of a mechanical load has to be put on the motor.  What is clearly implied is that just by virtue of the fact that the motor is spinning, then you are getting a "mechanical output."

You have probably looked at just as many Bedini clips as me and you have heard that statement many times.  Nobody ever mentions that they have to put a mechanical load on the rotating shaft of the motor and they seem to believe that they are getting mechanical output just by staring at it spinning.

I stand by what I said, the mechanical output of a Bedini motor in a standard setup with a charging battery is zero.

In addition, in the 10-coiler infomercial, Bedini mentions the small pick-up coil to power the LEDs that light up the clear plastic frame and he says something like "the free mechanical is powering the LEDs."  That's not magical free energy that is coming off the spinning rotor of the 10-coiler.  That energy comes from the battery.  I even give suggestions on how to test that for yourself.

What are your thoughts on Bedini's statement about the "29% mechanical power?"  What do you think that means?  How did he arrive at that number?  I would like to know because I can't figure that one out.

MileHigh
   
Group: Guest
Ren:

Just to emphasize my point, let's look at the Bedini quote again:

Quote
but please do not leave out
> the wheel rotation in you calculation, mechanical power is equal to
> work done, its a figure of 29% so what kind of machine have you built?

Do you sense that John Bedini is implying that you have to add a mechanical load?  I sure as hell don't.  He is clearly stating that the "wheel rotation" is "work done" a.k.a. "mechanical power (out)."

Read it again, what do you think he is implying?

MileHigh
   
Group: Guest
Ren:

Just to emphasize my point, let's look at the Bedini quote again:

Do you sense that John Bedini is implying that you have to add a mechanical load?  I sure as hell don't.  He is clearly stating that the "wheel rotation" is "work done" a.k.a. "mechanical power (out)."

Read it again, what do you think he is implying?

MileHigh

I agree with what you say , the only thing that come in mind is Bedini loaded it until it come close to stop and calculated a ratio with the input vs output from the mechanical work done, just by placing coils around to act like a generator, its easy to know but we need to factor the result in the charging battery too , if 50% less power go to the charging battery its not 29% mechanical power anymore but if the charging battery charge at the same rate joule in/joule out (always in comparaison with the input power with the mecanical load vs charging power output and input power with no mecanical load vs charging power output), that can confirm the 29% mecanical power (torque meter can be used too or coils setup to act like a generator).

Best Regards,
IceStorm

   
Group: Guest
Ren:

Just to emphasize my point, let's look at the Bedini quote again:

Do you sense that John Bedini is implying that you have to add a mechanical load?  I sure as hell don't.  He is clearly stating that the "wheel rotation" is "work done" a.k.a. "mechanical power (out)."

Read it again, what do you think he is implying?

MileHigh

I dont have to wonder what he is applying. Attach a fan blade to the shaft, rewire a fan if you cant be bothered building one from scratch, hell, you are overcoming magnetic cogging, friction in the bearing and air resistance without it anyway. All of which require SOME mechanical. Im not here to argue percentages, nor am I here to pick Johns sentences apart, and wonder what he is implying. Im here to argue with your blunt statement that the mechanical power is ZERO. And that the rotor is the equivalent of a resistor, heating up and "pouring" energy down the sink. Which is ridiculous, my circuits dont heat up at all (nor does my rotor) even at higher voltages.

No matter how small the mechanical power is, it is still present. Otherwise, by your defiiniton it would continue to spin indefinately, as it would have to see no mechanical resistances to qualify as having no mechanical power. If it overcomes anything that seeks to bring it to a halt then by definition it has to have mechanical power of some sort.

I guess we will just have to agree to disagree here. Ive already butted my head up against the wall trying to explain what I thought was very clear to see.



   
Group: Guest
Ren:

As I have explained, the most basic energy analysis of a Bedini motor is that it produces heat and charging, giving you a power pie-chart  cut into two pieces.  If you add a fan to the rotor shaft, then it produces heat and charging and drives a mechanical load, and the power pie-chart is split three ways.

Here is an analogy:  You are in an automotive research center in the engine design department.  You are looking at a new engine design and the engine is idling and connected to a transmission.  The whole assembly is sitting on a mounting in the middle of the lab.  Otherwise there is nothing connected to the transmission.  You ask the engineer, "What's the current mechanical output in horsepower?"  He will say to you, "Currently the mechanical output is zero.  When you see the engine idling like that, all of the power output is just overhead to keep the engine running.  However, we have a special automotive pony brake that we connect to the output of the transmission to run our tests.  Only when we connect up the pony brake to the transmission output shaft can we start to measure usable horsepower from the motor."

So in the case of the Bedini motor, all of the torque produced when the rotor is turning at its normal speed is being used to overcome bearing and air friction to maintain a certain RPM.  They are in balance and cancel each other out so that the net output torque is zero.  Let's suppose that the rotational speed at this balance point is 1000 RPM.  Then you attach a fan to the shaft.  Now you are getting usable torque from the Bedini motor and it is exporting mechanical power to outside world.  You also notice that the rotational speed dropped to 940 RPM when you attached the fan.

What this means is that you can plot a performance curve for your Bedini motor.  Let's say the vertical axis is RPMs and the horizontal axis is torque.  The curve starts on the upper left at zero torque and maximum RPM.  As you move to the right and the torque starts to increase the RPMs start to drop.  So you end up plotting a line that starts at the upper left of the graph and slants downwards to the lower right.  The lower right is maximum torque and minimum RPM.  As you might imagine, you could also plot other types of performance curves to describe the behaviour of the motor.

Quote
my circuits dont heat up at all (nor does my rotor) even at higher voltages.

The key thing is to be aware that your circuits are producing heat, even though sometimes it might not be perceptible, it is still there.  Same thing for the spinning rotor agitating the air.  You know that it takes some torque times an angular velocity to overcome the air friction on the spinning rotor.  That mechanical power has to go somewhere and it becomes heat power that heats up the air.

You have to keep in mind that all of the generation of heat power is by definition power that you have lost, because it serves no useful purpose for your application.   You put a Bedini motor in a "black box" and run it for one hour, and you put a solid-state Bedini setup in another black box and run it for an hour.  After the test is over in both cases you end up with excess heat and a charged battery inside each black box.  You can't tell which black box contained the Bedini motor vs. the black box that contained the solid-state Bedini.  That's just another way of illustrating to you that the mechanical output is zero from a standard Bedini motor setup.

MileHigh

« Last Edit: 2010-09-22, 16:05:18 by MileHigh »
   
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"Let's suppose that the rotational speed at this balance point is 1000 RPM.  Then you attach a fan to the shaft.  
Now you are getting usable torque from the Bedini motor and it is exporting mechanical power to outside world.  
You also notice that the rotational speed dropped to 940 RPM when you attached the fan."

@ MileHigh...
True but this same thing is happening on a fossilfuel engine. Only thing is that the engine automalticly adjust itself by using more fuel to keep the desire RPM.
If u should take that away and load that engine heavy enough it will slow down and even stop spinning. That sound like what a bedini is doing.
The thing is people tend to make a comparison between a Bedini and conventional fossilfeul engine.
Can u drive a Big dump truck with a small 1liter engine from a suzuki alto? That is like the comparison of a bedini with a combustion engine.
If we approche these devices this way we will not get the job done.
Example, the setup i have now.
Someone ask me if i can run my house on it, well i can if i use a combustion engine to run it. But thats not the point or the approche here.
I want to drive it with a pulse motor so the approche will/must be different.
I feel like we humans have gone stuck and we are affraid to take a step backwards so we can improve in the future.

U can load the shaft of a bedini but one have to take several things in considiration otherwise it wont work.
U can even load the shaft without decreasing its primal RPM.
A hint, give alittle and than take alittle though we are programmed to only take as much as possible.
And one just have to find a good balance between the system and its load. (not like a truck example above)

About the heat, there maybe some heat but not much depending on the setup.
The rotor going through the air does produce some heat by its friction between the air but at the same time it produces a nice wind fortex that cools off the bearing, itself and the whole unit.
Putting it in a box will definately kill it...haha...put a combustion engine or any kind of engine in a box and see what happends...haha

Oke i hope that makes some sence but than again i may be a complete fool at this though :P...haha

@ Ren...
Are u the person i think u are...if so...how u been m8.
And hows your ceilling, i do hope u havent got more holes in it by now...haha...inductive launcher, housten we got a problem.
oke, abit offtopic.

Modify 6 times for typos... :( ::)
   
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Hi BiDaDiKuNuKu:

Quote
True but this same thing is happening on a fossilfuel engine. Only thing is that the engine automalticly adjust itself by using more fuel to keep the desire RPM.

Yes I agree with you.  You could easily have a speed governor connected to the Bedini motor so when you add the fan to it, the increased load is compensated for by increasing the power supply voltage so that the Bedini motor maintains a constant RPM.  In engineering terms it's called a "negative feedback loop" and they are built into car engines but they aren't normally built into Bedni motors.  The main point is still about where your battery power is going and how much of that battery power is doing useful work for you.

Here is another example from the automotive world:  Some cars have "trip meters" and they tell you your miles per gallon (or liters per 100 Km) in real time as you drive your car.  The idea is to encourage motorists to drive in a way to increase their fuel economy.  However, when you come to a red light and stop, then look at your miles per gallon, the trip meter display says that you are getting zero miles per gallon (or infinity liters per 100 Km!).

The whole point of this exercise is to make people aware that when you say "you also have to include the mechanical output" for a Bedini motor when discussing if it might be an over unity device, that this is a false statement.  I have seen John Bedini state it, Rick Friedrich state it, and perhaps a dozen other experimenters state it.  Unfortunately, they are all incorrect.  It's all about the truth.  It's not "mechanical power," it's waste heat power.

When you say, "you also have to include the mechanical output" to someone, you are planting a false idea in their head.  You are setting up a knowledge structure and a value proposition about a Bedini motor in the minds of the enthusiasts that is not true.  This thread is about helping experimenters find the truth about their Bedini motors.

If you want to connect something mechanical to a Bedini motor and measure the mechanical output power, that's fine and go for it.  However, when you do that, if you still want to have the same amount of charge power going to the charging battery, then you have to either adjust the trimpot for the base resistor and increase the power consumption of the motor, or perhaps increase the source voltage for the Bedini motor.

So what this means is this:  If you want to have the same charging power from your Bedini motor after you add a mechanical load, then you have no choice but to increase the power that you are supplying to the Bedini motor.  In other words, any mechanical output that you want to get from a Bedini motor is NOT FREE.  Bedini says this and it is simply not true.  You have to pay for it.  This can all be proved through experimentation.  There are two or three postings in the middle of this thread devoted to testing if the mechanical output from the Bedini motor is free or not free.  If someone was determined to check this for themselves it can be done.

MileHigh
« Last Edit: 2010-09-22, 17:20:23 by MileHigh »
   
Group: Guest
Quote
"Let's suppose that the rotational speed at this balance point is 1000 RPM.  Then you attach a fan to the shaft. 
Now you are getting usable torque from the Bedini motor and it is exporting mechanical power to outside world. 
You also notice that the rotational speed dropped to 940 RPM when you attached the fan."

I got it. Usable is the key here. While just spinning with no load the load becomes the friction, air movement, heat and so on for such X of input of power. Tending to zero net output as it all balances and no MORE USABLE power is extracted.

So when one say "K mechanical power is free" it should be qualified to what Y more of cost it will come. If Y is not defined or implied zero so the statement becomes false.

Did I got it correct MileHigh?

Fausto.
   
Group: Guest
@ MileHigh,
I got your point and i`ve did not stated that bedini is overunity...maybe bedini did but not me... ;D

"When you say, "you also have to include the mechanical output" to someone, you are planting a false idea in their head.  You are setting up a knowledge structure and a value proposition about a Bedini motor in the minds of the enthusiasts that is not true.  This thread is about helping experimenters find the truth about their Bedini motors."

I did not want or intend to plant anything in the minds of the bedini enthusiast. ??? :o
I only said what that i can load the shaft on my pulse motor without my pulse motor using more energy. ;)
But one must take some things in considiration. But than again my setup isnt like a bedini setup.
Bedini is a pulsed motor and so is mine but with some twist here and there.

But if one is using a normal bedini system that would be a different story "i think"
I think i`ll leave this thread here for the real bedini expert since my setup isnt a bedini but still its a pulse motor.

Have fun  ;D
   
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Fausto:

Yes you got it.  Any mechanical power output from a Bedini motor has to be usable torque that actually does something tangible.  It could be driving a fan to blow air, or driving a water pump, etc.

BiDaDiKuNuKu:

Quote
When you say, "you also have to include the mechanical output"...

The first "you" in my statement does not refer to you personally.  It is more like saying, "When they say..."  I think in one of your clips you state that you speak four languages, which is amazing.  I am sorry if there was any confusion.

Quote
I only said what that i can load the shaft on my pulse motor without my pulse motor using more energy.

Yes, this is a property of a pulse motor.  In many cases when you (not you personally) put a load on the shaft of a Bedini motor the power consumption of the motor actually goes down.  This is another case where many Bedini enthusiasts misunderstand what they are observing.  What's happening is when you put the mechanical load on the Bedini motor, it slows down.  Since it slows down it is consuming less pulses of current per second from the source battery, and therefore the power consumption goes down.

Many enthusiasts see the power consumption drop when they put a mechanical load on their Bedini motors and think that this is even an indication of free energy.  They think that the power consumption should increase when they add the mechanical load.  What is happening is that the "average impedance" of the motor changes from the point of view of the drive battery.  If the average impedance increases, then the power consumption will go down.  If the average impedance decreases, then the power consumption will go up.

Depending on how the specific Bedini pulse motor actually works, the power consumption could increase or decrease when you put a load on it.  This has to be evaluated on a case by case basis.  The reason for this is that the lower pulse rate would normally indicate lower average current consumption.  At the same time the lower pulse rate may result in the transistor being on for a longer amount of time for each pulse, and that might increase the average current consumption.  I am pretty sure that I discuss this in more detail earlier on in the thread.

If you leave the thread as a contributor, I hope that you will continue to read it anyway.  I hope that some of the information has been helpful to you.

MileHigh
   
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@ MileHigh...
"you" in general way of speaking, got it m8... ;)
About me speaking 4 languages, sounds cool but afhther years of speaking all 4 some what at the same time with different persons my brain tends to fuse them like its trying to make a new language...one can hear that in my vids to...anyway ;D

About me contributing on this thread, i dont know about that.
But i`ll hang around and read here and there and asking some stuff even though i dont like to ask much.
Like i stated earlier, i am no expert on these stuff but i still give it a go.
That might be my advantage... :P

About the bedini decreasing amp draw when one put a load on it.
In all the stuff i made whenever i put a load on it, just my hand, light bulp, radio or whatever the case with me the amp draw always goes up.
But than again i never have made a bedini like bedini say one must.
I`ve used the circuit but my coils, rotor and magnetes arrangements where and still are different... ;D

Peace!
VZ2DAY
   
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@MileHigh,

this thread is really good. I have been reading it many times to get the concepts correct in my mind.

So concerning an inductor, I understand, as you said, its initial current discharge is the same as its initial current right before the "discharge" phase.

So is it correct to assume it is impossible to have a coil in "charged and disconnected" state? Since, as soon as one try to disconnect a coil it will discharge? While a cap you can have one disconnected from anything and still hold the charge?

Another questions is: when you say that the amount of energy leaving the primary is the energy that is going to the secondary battery via the coil, (less losses) since it will have to discharge into the secondary, the only special thing is that the coil will "adjust" itself to the "load" and change the voltage and time of discharge? So that's, may be, what Bedini means when he mentions impedance matching and you say that there is NO impedance matching because the coil does it?

Another question: when they propose the one ohm resistor test as a proof that there is "no current" going into the battery would be equivalent to the following process:

coil charges with a certain amount of energy, lets say 10watts from whatever current and voltage that it received from the primary. If the load is the one ohm resistor, so the voltage cross the resistor will be very low and current very low but a long pulse? But still the same amount of power delivered in that "time it takes to discharge", again less or almost equal to 100% of initial power, in this case 10watts? So they see a small current and claim the battery being charged with "no current"?

If this is correct, should not the resistor still get as hot as 10 watts equivalent?

many thanks,

Fausto.



   
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Hi Fausto,

I am really glad that you are enjoying the thread and learning.  That's the reason I made it.

Quote
So is it correct to assume it is impossible to have a coil in "charged and disconnected" state? Since, as soon as one try to disconnect a coil it will discharge? While a cap you can have one disconnected from anything and still hold the charge?

Yes you are correct, the moment you disconnect the coil it will instantly discharge its stored energy.  Think about the fact that the coil stores the battery energy because of the flowing current.  So if you have no current flowing through the coil then it is impossible for any battery energy to be stored in the coil.

With a coil, in theory you can keep it charged but you have to connect the two terminals of the coil together so that current continues to flow through the coil.  In other words you have to short-circuit it.  Of course we know that there is resistance in the wires of the coil, so if you short-circuit the two terminals eventually the current will stop flowing through the coil.

For example, you could connect a coil to a battery, wait a few seconds, and then close a switch that shorts the two terminals of the coil together.  Naturally this will also short the two terminals of the battery together, so immediately after you close the switch you have to disconnect the battery.  The coil will then store the energy that came from the battery because current will continue flowing through the coil.

Quote
Another questions is: when you say that the amount of energy leaving the primary is the energy that is going to the secondary battery via the coil, (less losses) since it will have to discharge into the secondary, the only special thing is that the coil will "adjust" itself to the "load" and change the voltage and time of discharge? So that's, may be, what Bedini means when he mentions impedance matching and you say that there is NO impedance matching because the coil does it?

Yes the coil will adjust its voltage depending on the load.  However, there is no impedance matching going on.  Let's look at a similar scenario with a capacitor.  You have a 1000 uF capacitor charged to 10 volts.  You connect a 100-ohm resistor across the capacitor and discharge it.  You then do the same thing again but this time you use a 1K-ohm resistor.  In both cases the capacitor gets discharged.  With the capacitor, does it sound like impedance matching to you?  I am assuming the answer is "no."  It's the same thing for an inductor.

Impedance matching means that you want the internal impedance of the power source (say like a 1.5-volt battery) to be the same as the impedance of the load.  So if a 1.5-volt alkaline battery has an internal impedance of 0.1 ohms, then you have to put a 0.1-ohm load resistor across the battery to get the maximum power transfer rate.  Discharging inductors and capacitors are completely different from the alkaline battery example.

If John Bedini or others make the false claim again that the Bedini motor output is doing "impedance matching" then just ask them what the value of the impedance match is.  I don't think they will be able to give you an answer and they will try to deflect the question away.  Look at my example above, I give you a clear example where the value of the impedance match is 0.1 ohms.

The short answer is that both an inductor and a capacitor can discharge all of their stored energy for many different load resistances.  "Impedance matching" implies that there is a single correct load resistance.  Therefore by definition there is no impedance matching associated with a Bedini motor drive coil output.

Quote
Another question: when they propose the one ohm resistor test as a proof that there is "no current" going into the battery would be equivalent to the following process:

coil charges with a certain amount of energy, lets say 10watts from whatever current and voltage that it received from the primary. If the load is the one ohm resistor, so the voltage cross the resistor will be very low and current very low but a long pulse? But still the same amount of power delivered in that "time it takes to discharge", again less or almost equal to 100% of initial power, in this case 10watts? So they see a small current and claim the battery being charged with "no current"?

If this is correct, should not the resistor still get as hot as 10 watts equivalent?

You are confusing power and energy.  Power is in watts and energy is in joules.  Coils store energy in joules.

For a one ohm resistor, the pulse will be a long pulse.  The initial current will be whatever current was flowing through the coil the instant before the transistor switched off.

Fausto, your answer has already been answered several times in in the thread.

Instead, let me request that you to do two things:

1.  Review power and energy concepts several times so that you can reread your last question above and see the mistakes.
2.  Do three tests with your Bedini motor where you replace the charging battery with 1-ohm, 10-ohm, and 100-ohm resistors.  Look at the voltage across the different resistors with your scope when the motor is running.  You know the resistance, so when you look at the voltage you also know the current.  What can you conclude about the power and energy in each case?

Thanks,

MileHigh
« Last Edit: 2010-09-28, 03:55:25 by MileHigh »
   
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Fausto:

Now I am going to give you a short answer to a short version of your last question:

Quote
Another question: when they propose the one ohm resistor test as a proof that there is "no current" going into the battery

The short answer is that when someone says that a Bedini motor charges a battery with "voltage but almost no current" they are completely wrong.

The discharging drive coil ALWAYS outputs a spike of current and a corresponding voltage spike.  Current has to flow, and the resultant voltage is variable.  There are no exceptions to this because we are dealing with a discharging inductor, and a discharging inductor is a current source.

The simple truth Fausto is that the vast majority of people that experiment with Bedini motors are not aware of these facts.  If you do the simple experiment with the three different load resistors it should help you understand this.

MileHigh
   
Group: Guest
@MileHigh,

perfect. Excellent. That is what I thought. I do understand your arguments and yes I think I do understand the difference between watts and joules. I really meat 10 watts as an instance in time of the energy not the long running accumulation of it in joules. Just as an example of transfer of power/energy in a timeless fashion.

Thank you. Your answers are very valuable. I don't understand why people would be concerned with your arguments. If Bedini claims are not to the dot or even incorrect I think there is nothing wrong to understand both camps and adjust oneself to the reality in front on them.

I am loving this learning experience and I am still playing with this motors. Nothing better than things like this to get involved. After all we are trying to change the world into a better place.

Fausto.
   
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Posts: 2689
@milehigh
Quote
The short answer is that when someone says that a Bedini motor charges a battery with "voltage but almost no current" they are completely wrong.
LOL, are you still trying to confuse everyone here?, a discharging coil (an inductive discharge) will raise it's voltage in proportion to to the resistance/impedence it encounters and raise it's voltage in proportion to the rate of change of the switching mechanism which is a form of resistance. If the resistance is infinite to an inductive discharge then the voltage is infinite ( E=IR) and as the energy cannot change the current (I) must approach zero. Now if the discharge voltage spikes to hundreds of volts and the energy is constant then obviously there can be almost no current or do you not agree with ohms law?. As well if two conductors attached to a battery have a potential difference across them higher than the battery potential such as in an inductive discharge then this is defined as a charging action thus the battery must charge, Im really not sure which part of this you do not understand.

Quote
The discharging drive coil ALWAYS outputs a spike of current and a corresponding voltage spike.  Current has to flow, and the resultant voltage is variable.  There are no exceptions to this because we are dealing with a discharging inductor, and a discharging inductor is a current source.
Once the circuit is disrupted(opened) then the coil becomes a source in itself as the collapsing magnetic field is cutting the conductors of the coil, now how do you think the coil became a source? It became a source because as Ampere stated the field can be considered as "seperate" from the force which created it but all forces must translate back to the source. When the coil current is disrupted the coil becomes independent of the source because the magnetic field is independent of the source and can generate any voltage proportional to the resistance it encounters in which case it can become a current or a voltage source. Did the collapsing field induce a current? No, the collapsing field induced a potential difference by cutting the coil conductors and if the coil is in a circuit then this potential difference will lead to an electric current thus the voltage came first, it is a voltage source first and the current is a result of the voltage discharging itself into a circuit.

Quote
Yes the coil will adjust its voltage depending on the load.  However, there is no impedance matching going on.  Let's look at a similar scenario with a capacitor.  You have a 1000 uF capacitor charged to 10 volts.  You connect a 100-ohm resistor across the capacitor and discharge it.  You then do the same thing again but this time you use a 1K-ohm resistor.  In both cases the capacitor gets discharged.  With the capacitor, does it sound like impedance matching to you?  I am assuming the answer is "no."  It's the same thing for an inductor.
Right, can a capacitor raise it's voltage in proportion to the resistance it encounters? Well no it cannot thus it would seem obvious that a discharging capacitor has little in common with a discharging inductance.

Quote
The short answer is that both an inductor and a capacitor can discharge all of their stored energy for many different load resistances.  "Impedance matching" implies that there is a single correct load resistance.  Therefore by definition there is no impedance matching associated with a Bedini motor drive coil output.
The first problem here is that a battery is not a resistor and if you put them side by side most everyone might say--"hey they are not the same thing" and these people would be correct in my opinion. A battery will raise it's voltage in proportion to the voltage of the source charging it and we call this action "charging" where a resistor will not because it is not "charging" it is dissipating energy as heat, they are not the same thing. One could also ask why all the new battery chargers use boost converters exclusively where 20 years ago they used transformers, one word --efficiency. I wonder why Mr.Bedini was pulse charging his batteries with an inductive discharge 20 years ago like we are today when all the EE's were using lossy transformers, did he know something they didn't, lol? I also have to wonder why Tesla started using impulse currents(inductive discharge) exclusivelty almost 100 years ago, one word---efficiency. The fact of the matter is that a battery will raise it's voltage and resistance in response to the action of "charging" and the inductive discharge will raise it's voltage in proportion to the batteries rise in resistance as it must, does this sound like impedance matching?. My battery chargers and the batteries they charge run cool, not hot, not warm---cool and I do not believe anyone should need to be a rocket scientist to understand why this is the most efficient way to charge batteries.

Regards
AC


---------------------------
Comprehend and Copy Nature... Viktor Schauberger

“The first principle is that you must not fool yourself and you are the easiest person to fool.”― Richard P. Feynman
   
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Posts: 2689
double post


---------------------------
Comprehend and Copy Nature... Viktor Schauberger

“The first principle is that you must not fool yourself and you are the easiest person to fool.”― Richard P. Feynman
   
Group: Guest
AC:

Quote
LOL, are you still trying to confuse everyone here?

What are you trying to prove AC?  What a great way to start off a posting.  What is this business with the fake condescending attitude and the swagger?

At least half a dozen times in the past you have tried to put forward strange propositions and I rebutted them.  You didn't come back to try to argue your points further.  The best that you could do was to go off on ridiculous tangents as a means to reinforce your points.  Then you crumbled and faded away every time.  In your most recent posting, besides the attitude, you clearly show that you know some stuff, and at the same time you clearly show your technical limitations.

The person that is confused is you, AC.  Almost everything you said in your last posting should be ignored because it is wrong.

MileHigh
« Last Edit: 2010-09-28, 12:42:44 by MileHigh »
   
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