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Author Topic: Flynn's Parallel Path  (Read 58956 times)

jake

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Re: Flynn's Parallel Path
« Reply #15 on: May 25, 2006, 03:24:40 AM »
Quote
Now, does the flynn switch require more power when the motor is under load?
Or is the power supplied to it constant, unchanging regardless of motor load?

Username_1,

Flynn type designs purportedly do not exhibit the typical Torque = K*I relationship that a typical motor does.

The motor current is said to increase slightly as the motor first starts to load up, but quickly flattens out.

The motors are said to run extremely cool (evidence of the touted "high efficiency"), and are supposedly virtually impossible to "burn up", unless you inadvertantly keep a coil energized when the motor is not running.

The coils on Flynn motors supposedly exhibit a "kick" back from the coils that gets more pronounced as the motors load up.

It seems like Flynn stuff is producing some very interesting test results, and is being reproduced by others.

Interestingly, I don't think Flynn ever touted his motor to be "overunity".  He just sold it as being very efficient and let it speak for itself.

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Re: Flynn's Parallel Path
« Reply #15 on: May 25, 2006, 03:24:40 AM »

jake

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Re: Flynn's Parallel Path
« Reply #16 on: May 25, 2006, 02:02:39 PM »
I don't think anyone says that the Flynn technology is producing more output power than the electrical power it takes to run it. (it may be, but I don't think they are saying that openly)

I don't have the STAIF document yet, but I also heard that it showed test results that were "overunity" in a portion of the operating curve.

I think the Flynn strategy was clever - keep the claims modest and let the hardware do the talking.

Offline Elvis Oswald

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Re: Flynn's Parallel Path
« Reply #17 on: May 25, 2006, 06:59:30 PM »
I obviously didn't make myself clear in the earlier post.   :-\

In regards to the Flynn tech... I was not talking about motors.

You are all talking about running a generator with the motor... and that may or may not yield overunity... but that's not the best way to use Flynn tech to acheive overuntiy.  With all due respect, doing it like that is like going around your ankles to wipe you ass. ;)

I'm talking about a transformer designed to multiply power.  No moving parts - just a transformer.

Using the parallel path setup, the magnetic force is amplified, and so then should be the current induced in a secondary coil.  From the looks of the test I saw... that force should be just under 2x.

So all that is to be done is to wire it up to run itself.  And this is where the questions come in... and what I was trying to suggest discussing in my previous post...

To start... I have to correct something someone said - Input power does matter.  The parallel path only manifests in a small window of voltage.  So, to power itself, the input power has to remain steady.
But you wouldn't want a string of 100 of these things to step up from some small amount of power... so you have to come up with something to collect the power.

Maybe connect one to something like the voltage multiplier on a microwave.  You might be able to make a small unit that you could start off by placing a magnet on the side.







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Re: Flynn's Parallel Path
« Reply #17 on: May 25, 2006, 06:59:30 PM »
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Offline Elvis Oswald

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Re: Flynn's Parallel Path
« Reply #18 on: May 26, 2006, 05:13:36 PM »
Just a note on the '3.47 times the force'  - compared to what?? 

I'd prefer to reference the difference between the setup with and without the parallel path effect... and that would be about 2 times.
The setup with the magnets - but without the coils lifts 'X', the electromagnet lifts 'Y' - by adding the coils to the setup with the magnets... you would expect to lift X+Y... but it actually lifts X+Y+Y.  It's actually pretty close to doubling the force of the current applied.

Offline Elvis Oswald

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Re: Flynn's Parallel Path
« Reply #19 on: May 26, 2006, 08:47:11 PM »
I would consider the lifting of the weights to be a measurement of force.  But even so - there is an increase that would equate to an increase in power to a secondary circuit.
And of course the power in would have to remain the same.

If you take the simple experiment (from one of those links) - the one with the steel straps and the strapping tape - and use an A/C current (of the voltage*amps that correspond to the pp effect) you should notice an increase in power.  I'd wire the two secondaries together, series... checking polarity twice... and you should see more power out than in.

If that works... maybe a HF A/C input would help multiply these small voltages faster?  Maybe increase the frequency with load?

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Re: Flynn's Parallel Path
« Reply #19 on: May 26, 2006, 08:47:11 PM »
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jake

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Re: Flynn's Parallel Path
« Reply #20 on: May 26, 2006, 11:31:44 PM »
If you analyze the "Simple Magnetic Force Multiplication Experiment", using the principles of magnetism spelled out by Flynn, the results of the experiment are not surprising or astounding.

The formula for force in magnetisim (per the Flynn site, and presumably common accepted knowledge) dictates that the attractive force increases with the square of the flux.

Now if we look at the forces in the experiment and relate flux values to them we get the following:

Lets assume 421 grams = 1 unit flux
If we add another unit of flux (one more permanent magnet), per the above formula, we should expect exactly 4 times the lifting force (2 times 2, or 2 squared) which would be 1684 grams.

The actual experiment result was 1721 grams, which would indicate that the second magnet was just slightly stronger than the first magnet.

So far, what has happened is exactly what would be expeced.  We doubled the flux, and quadrupled the force.  If there is anything telling at this point, it is that 100% of the flux from the top magnet is acting on the bar across the bottom. (without any "steering")

Now, to go from the 1721 grams of force in the second diagram to the 3845 grams in the third figure would require:
(3845/1721)^0.5 = 1.494 times the flux that we have in the second diagram.

In the second diagram (1721 grams) 2.02 units flux are present: (1721/421)^0.5=2.02

This means that the third diagram would have 2.02 * 1.49 units of flux, which equals 3.02 units of flux.  Can also be computed as a ratio of diagram 1: (3845/421)^0.5 = 3.02 as a back check on the figure.

Since we have 2.02 units of flux in the second diagram, and 3.02 in the third diagram, we should have to add 1 unit of flux to arrive at the 3.02 units of flux that would be required to give us the 3845 grams of lift.

Everyone follow me so far?

This means that the electric windings are adding exactly the equivalent of 1 unit of flux to the system, which would be expected to give the zero force at the top, effectively "switching off" the top magnet, right?

Now we go to the last figure.  This figure shows that we are applying electrical current that would produce 1091 grams of force, which would require (1049/421)^0.5 = 1.58 units of flux.

What all this adds up to is we are applying electrical current to the circuit that equates to 1.58 units of flux, as proven by the right diagram in the experiment.  The net gain we are getting from applying current that should produce 1.58 units of flux is only 1.0 units of flux in the third diagram.

The upshot is, we are applying enough current to the circuit to expect a much larger lifting force that we actually get.  If the 1.58 units of flux being generated in the circuit on the left were added to the 2 units of flux in the second circuit, we should see 3.58 units of flux, or 5396 grams of force in diagram 3.

If we take the information from this experiment and analyze what would happen if we attempted to use it to make a "super transformer", we can also predict what would happen.

There is no arguing that we have fundamentally 3 uints of flux in diagram 3, with the current applied.  By putting a bar across the top we can prove that there is no flux available at the top of the circuit.  (see Flynn documents showing zero force on the top bar).  Now, if we reversed the polarity, we would have 3 units of flux at the top, and 0 at the bottom.

If we put a bar across each end, with a coil around each bar, can now create a current in the coils, which is based upon the change in flux passing through the coils.  Each coil will oscillate between 0 units of flux and 3 units of flux as we alternate the current to the coils.  This gives us a delta of 3 units of flux in each coil, by applying electricity that would create 1.6 units of flux.

Now lets take the magnets out of the circuit.  We now have figure 4, but add the coils at each end of the circuit.  This circuit differs in that the magnetic flux will completely reverse as we alternate the current to the input coils.  Since it alternates (because the permanent magnets are not there to "redirect" the flux), we will oscillate from +1.6 units of flux to -1.6 units of flux (complete flux reversal at both ends).  This means that the coils will see a delta of 3.2 units of flux without the magnets in the circuit.

What this means is the permanent magnets actually make the transformer less effective by the ratio of 3.2 to 3.0, or about 93.5 percent.

Thus, I predict that if someone makes the circuit shown and adds the coils at each end and makes it so that it can have the magnets put in and taken out, that the transformer would work about 6.5% better by leaving the magnets out.

If you look at what has to happen in the circuit, all of the results make sense.  To get from 2 units of flux to 3 units of flux requires that the electric coils drive 1 unit of extra flux through the top permanent magnet to get it around to the bottom of the circuit.  Since the magnet is already providing 1 unit of flux, this effectively doubles the flux that must pass through the upper magnet.  This probably explains why they warn that Nib magnets are not good for this experiment.

I'm not saying all this to imply that the Flynn circuit is useless, but to point out that it is probably not magic.  Upon analysis the circuit works just the way it should.


Offline Elvis Oswald

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Re: Flynn's Parallel Path
« Reply #21 on: May 27, 2006, 04:02:56 AM »
The power to generate 1.6 units of flux will only produce 1.6 units... that's why it's called "the power to generate 1.6 units of flux.  ;)  That's the first thing you're missing.  In a regular transformer... with equal windings on both coils.... power out = power in (best case)
So your estimate of power out of a regular transformer is wrong.

If you are correct on the math about the transformer with pp... then the power out should be almost double.  And since that jives with my abstract calculations, I would say that we are looking at doubling power with a simple transformer.

Free Energy | searching for free energy and discussing free energy

Re: Flynn's Parallel Path
« Reply #21 on: May 27, 2006, 04:02:56 AM »
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jake

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Re: Flynn's Parallel Path
« Reply #22 on: May 27, 2006, 05:50:17 AM »
Elvis,

My math says that if you get 1.0 without the magnets, you will get 0.93 with the magnets, so my math doesn't jive with your abstract calculations.

I encourage you to build the device and test it.

The first thing you are missing is that the 1.6 units of flux completely reverses when you reverse the polarity on the input coils when the magnets are not in the circuit.  This gives a change of 3.2 units when the magnets are not present (+1.6 - (-)1.6 = 3.2) ;)  With the magnets in the circuit, the flux does not reverse through the secondary coils when you reverse the polarity of the input coils.  It simply varies from 0 to 3.0 for a change of 3.0.  This is a diminishing return by having the magnets present in the circuit.  (The transformer output voltage is proportional to the flux change through the output coils)

Offline Elvis Oswald

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Re: Flynn's Parallel Path
« Reply #23 on: May 27, 2006, 06:05:09 AM »
What you are still missing is that - IF you use your example, and the output is 3.2... then you applying enough power to provide 3.2...

So stick the 3.2 into your first calculation and see how it changes.  Because the input is the same setups... and the input is "enough power to generate 3.2"  And that's not what you used in your first example.  You used enough power to generate 1.6.
You should have said it was +.8 and -.8 to equal 1.6



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Re: Flynn's Parallel Path
« Reply #23 on: May 27, 2006, 06:05:09 AM »
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jake

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Re: Flynn's Parallel Path
« Reply #24 on: May 27, 2006, 03:18:48 PM »
Elvis,

Respectfully, build it and prove it to yourself.

Offline Elvis Oswald

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Re: Flynn's Parallel Path
« Reply #25 on: May 27, 2006, 07:09:37 PM »
I absolutely will build it.  But we are discussing your math... and your math is wrong.  And since this is a discussion of some importance in a public forum... I refuse to leave the faulty math in place.

When you were calculating, you used an input power that created 1.58 units of flux for the parallel path setup... and then you used input power that generates 3.2 units of flux in the second example.  So your answer is wrong because you are using two different inputs.

I would appreciate a correction.  But if you want to leave such an error in your post... that's fine as well.  But I will correct you for everyone else, otherwise people would see your conclusion and assume it's right.

So to others discussing parallel path - please note that Jake's conclusion is incorrect.  But it is easily corrected.

In the first calculation he uses the parallel path setup and uses the correct input power.  The result shows that output is 3.0 units from an input equal to 1.58 units.

The second calculation is where his error occurs.  Instead of using the same power in... he uses power equal to 3.2 units.
If he had used 1.58 units - like he did in the first equation, and like he should have... then the output would be 1.58.

So, after correcting for the error... we see that by Jake's calculations, a pp transformer would have a net gain in flux in the core over a standard transformer.

Will that transfer directly to more current in the secondary?

Free Energy | searching for free energy and discussing free energy

Re: Flynn's Parallel Path
« Reply #25 on: May 27, 2006, 07:09:37 PM »
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jake

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Re: Flynn's Parallel Path
« Reply #26 on: May 27, 2006, 10:42:47 PM »
Elvis,

I'm sorry I can't explain this in a way you understand, but I assure you my math is correct.  The 3.2 units is 1.6 plus 1.6.  To create a transformer you are going to have to reverse the polarity of the current to the coil.  The reveral of polarity causes the flux to completely reverse if the magnets are not in the circuit.  With the magnets in the circuit, the flux at the ends (where the output coils would be in the transformer) does not reverse directions when the electrical input to the coils is reversed.  It drops to zero.

When you have the magnets in the circuit and you apply the current one direction you are going to get 3 units of flux at the bottom and 0 units of flux at the top.  When you reverse the current to the coils (you will have to do this to use the device as a transformer) you will have 3 units of flux at the top and 0 units of flux at the bottom.  This gives you a Difference of 3 units of flux

Now take the magnets out of the circuit (like the 4th diagram in the flynn experiment.  You have 1.6 units of flux in the bottom and 1.6 units of flux in the top with current applied.  Now when you reverse the polarity, you have -1.6 (1.6 units in the opposite direction) units of flux at the top and at the bottom.  This gives you a Difference of 3.2 units of flux (to be exact, 1.58*2= 3.16).

Now, if you say you don't want to reverse the current in the circuit you will still have a similar problem.  You could just "pulse" the input to the coils.  What happens then is you have a change from 2 units of flux to 3 units of flux with the magnets in the circuit for a total difference of 1 unit of flux.  Taking the magnets out and pulsing the coils will give you a change from 0 to 1.6 for a total change of 1.6 units of flux.

Quote
3845 grams od force (according to flynns test page) alternating , compared to the no magnet conventional curcuit  which gives  1091 grams.

This is absolutely wrong.  The force will go from 3845 to 0 to 3845 to 0, etc.  (The top coil has zero force on it - "the Flynn effect")  Thus when the current is reversed the bottom coil will have 0 force.

The other problem you are having is not respecting the relationship of force to flux (that Flynn spells out).  Force is proportional to the square of the flux.  A transformer functions based upon the rate of change of flux through the coils.  You can't use flux and force interchangably.  Twice the force doesn't mean twice the flux, so it doesn't mean a transformer will put out twice the voltage.

Offline acp

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Re: Flynn's Parallel Path
« Reply #27 on: May 27, 2006, 11:31:58 PM »
Yes, Jake's calculations make sense.

At any rate this modification sounds awfully similar to the MEG, which doesn't work anyway.....

Offline dutchy1966

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Re: Flynn's Parallel Path
« Reply #28 on: May 28, 2006, 03:49:49 PM »
Hello all,

I've been following these discussions here from the sideline. The flynn effect has been proven already and has recently been bought by Boeing.
The reason why most people cant get the transformer/meg type device to work is because some important information is missing from the patent. (probably left out on purpose!). This concerns the serial switching of the control coil and the output coil. Why this is neccessary? Because else the current supplied to the control coil would simply leak away in the output coil and no flynn/overunity effect would appear at all.
Read the attachmant i've include, it explains it all.

regards,

Dutchy

Offline Drak

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Re: Flynn's Parallel Path
« Reply #29 on: May 28, 2006, 10:20:15 PM »
  I have recently replicated the device. It does work. Using two "C" batteries, "I" laminents from a transformer, 22 guage wire, and four of the rectangle magnets from radio shack (not the big rectangle magnets, but the ones with the hole in center of them). Two magnets on one end and two on the other. After wiring accordingly it works as specified. The bar falls off of one end while the force on the other is considerably stronger. I haven't done weight tests yet, but by judgment from the feel of the pull it is pretty outstanding.

  When the flux is at one end while the device is on, there does not need to be a bar at the nonflux end as stated on one of the other sites for it to work. On the non flux end there is NO FORCE whatsoever. With out the magnets in place both ends have a magnetic pull, but nothing close the the pull on the directed path with the magnets in place and the device switched on.
 
  I just finished building the device at the beginning of the week. I do not have any weights to measure the pull with so I think I will use water as weight to mesure with. My Coils are not the best and I didn't use the same magnets, laminent size, or gauge coil as stated in the demo, but it worked anyways. I'm currently in the process of borrowing a camcorder, or at least a digital camera with motion if anyone is interest in seeing it.

  Oh yes, its not a good thing to try and use neomagnets to replicate this because of the saturation point of the laminets (so I read). Heh, I couldn't sleep that first night after building it, my mind was racing too much. Any how, out for now.

 Drak

 

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