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Author Topic: Charging a coil with less energy and get huge BackEMF energy pulses  (Read 21202 times)

Offline hartiberlin

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Hi All,
I had another idea, how to get a maximum
Back EMF energy pulse from a coil.

Normally if we charge up a coil with
current, so it has magnetic energy,
we need W= 0.5x L x i^2 energy
for doing this and have to wait at least
about 3 to 5 x Tau=L/R timconstants,
until the current builds up in a coil,
if we apply a DC voltage, where L is the inductance of
the coil and R is the ohmic resistance of the coil.

For reference see:
http://www.1zu160.net/elektrik/spule.php

Now imagine, if we could switch the inductance
from Zero to 100 Henries immediately !

This can be easily done with a bifilar coil,
where one half of the bifilar coil could
be shorted out !

This way, we could first switch the DC supply
voltage across the bifilar coil first and
as there is no inductance , so L= 0 Henries.
the current will flow immediately only
depending from the ohmic law I= U/R
, so from the Supply voltage and the Ohmic resistance
of the coil.

Now imagine, we have 10 Volts as the supply voltage
and the bifilar coil has 1 Ohm, so we can immediately
run a current of 10 Amps into it,
without waiting any seconds until the current is build up.

Now, when we short out one part of the bifilar coil,
by a flip of a shortout-flip-switch so the one part of the
bifilar coil still works now as a normal coil having for
instance 100 Henries, now we suddenly have a 100 Henry coil,
in which flows 10 Amps !

Normaly to build up a current of 10 Amps in a 100 Henry
coil  with 1 Ohm ohmic resistance at 10 Volts
supply voltage will take about
5 x tau= 5 x L / R= 5 x 100 / 1= 500 Seconds.

So all in all, if we first have the bifilar coil,
which compensates its own magnet field and thus
builds up the current very fast in an instance in a few mikroseconds,
depending only on the still applyable stray inductance
and then shorting out one half of it, so we suddenly
have a coil with 100 Henries and 10 amps in it
and we needed only for instance 10 mikroseconds to
charge it up, then we only had used:
Energy= Voltage x amps x time= 10 Volts x 10 amps x 10 mikroseconds=
1 MilliWattseconds of energy to charge this
coil up.

But according to the coil laws we would
then have stored inside the now 100 Henry coil:
Coil-magnetic-energy = 0.5 x L x i^2= 0.5 x 100 x (10 amps)^2= 5000 Wattseconds !

So now we have a coil charged with 5000 Wattseconds magnetic
energy, which we could get back by opening the coil
and discharge it into a load via the backEMF,
but only have applied 1 MilliWattseconds of energy
to charge it up...


So where is my error ?
Can this really work, or will the coil voltage or the coil current jump somehow, when we shortout one part of the bifilar coil ?

Regards, Stefan.

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Offline gyulasun

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #1 on: January 15, 2007, 05:55:03 PM »
Hi Stefan,

I would like to see your schematics how you mean shorting out one half of a bifilar coil when you apply the 10V.  Use simple switch and coil symbols, no need for a full (for instance a MOSFET switch circuit) drawing of course.

If you returned to this idea because you have seen/made some progress since your TEP idea then I would surely like to hear it, see this link to Naudin's site:
http://jnaudin.free.fr/html/tep60sh.htm   
http://jnaudin.free.fr/html/tep61sh.htm
http://jnaudin.free.fr/html/tep61sht.htm

It is very pity Naudin did not include the most important thing: Did he find overunity or not? How much is the free energy he refers to with respect to the input energy? (at least I have not seen it reported, did he inform you back then?).

My main problem with the idea is that if you short circuit one half of the bifilar coil then there cannot be any self inductance of the other, not shorted coil part because of their mutual closeness. Just the same way if you test a conventional mains transformer and short circuit its secondary coil, then the primary coil loses its self inductance (and heats up rapidly).

So if you returned to this idea due to some progress since then, please tell me what it is. 

Thanks,
Gyula

Offline gezgin

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #2 on: January 17, 2007, 03:57:07 PM »
It can be test, but 100H big coil isnt it?
if bifiliar coil L = 0, giving energy to coil zero or near zero or ?

I want to test this 97's circuit but how much henry must be bifiliar coil,
i have 5 mH coil but it isnt enought for test?

and another patented idea : parralell coils back emf device.

sorry my english

thanks

ilhan

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #2 on: January 17, 2007, 03:57:07 PM »
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Offline hartiberlin

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #3 on: January 18, 2007, 07:38:49 AM »
Hi Gyula,
you are right,
this was it, what I wanted to try...
Unfortunately when we did this in 1997 together with Dieter
Bauer, we had some input power coupling into the circuit
via the base of the transistor,
when I remember correctly and we had no time to
retry and set it up correctly...and retest it.

Indeed here is a strange thing at:

http://jnaudin.free.fr/html/tep61sht.htm

Why is then there at all a BackEMF coming out of this circuit from the Bifilar
coil at R3 ?

Is the voltage U(R3)
only from the stray inductance of the bifilar coil,
which is then a real inductance, small, but real, so this
stray inductance has stored this magnetic energy which is then
released, when the current inside the coils
is cut off ?


Can anyone explain this ?

Many thanks.

Regards, Stefan.

Offline gyulasun

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #4 on: January 18, 2007, 05:28:44 PM »

Indeed here is a strange thing at:

http://jnaudin.free.fr/html/tep61sht.htm

Why is then there at all a BackEMF coming out of this circuit from the Bifilar coil at R3 ?

Is the voltage U(R3) only from the stray inductance of the bifilar coil, which is then a real inductance, small, but real, so this stray inductance has stored this magnetic energy which is then released, when the current inside the coils is cut off ?

Can anyone explain this ?

Hi Stefan,

I discussed this circuit with a friend of mine and we agreed on the followings.

Let's suppose the forward voltage drop on D4 diode is 0.8V and the saturation voltage of Q1 transistor when switched on is 0.2V, this sums up to 1V, ok? 

Further, if we assume the battery is charged up to around a practical 13V value, then the useful voltage source is 13V-1V=12V, neglecting the internal resistance, ok?

Now let's consider Naudin scope shot on the U(AC) voltage first. It shows the voltage drop between the A and C endings of the series L1 and L2, ok? It is around 5.2V peak to peak, ok?
This 5.2Vpp voltage drop is due to the DC resistance of L1+L2 wires, ok?
Then the voltage drop on the R4 resistor should be 12V-5.2Vpp=6.8Vpp, ok?
This means a peak current of around 6.8V/1 Ohm=6.8A flowing out from the battery and this goes through all the way on L1, L2 and Q1 of course.

Now let's consider the U(R3) voltage, the back EMF voltage. When Q1 switches off, the A and C endings of L1 and L2 are floating, the only load is across L2 by the R3+D1 in series. The huge current suddenly terminating must induce a certain voltage in both coils due to
1) partly stray inductance indeed as you mentioned
2) the sudden collapse/disappearience of the magnetic field INSIDE the ferrit core
    because it can be considered as a magnetized core (like an electromagnet) during
    the time Q1 is switched on and ferrit cores do not normally keep magnetism, hence
    there must be a flux change in the core what induces the back EMF.
In fact this loaded back EMF value is bigger by 0.7V than the value shown in the scope shot because of the forward voltage drop of D1: so it makes around 1.4V (if I judge that pulse, U(R3)=0.7V also).

One more notice: the ferrit core can be saturated by this huge current and this makes the switch on/off process nonlinear, so the repeatabilty/reconstruction may involve inherent differences in the measured data.  This is nicely shown in Naudin's another test here:  http://jnaudin.free.fr/html/tep61a5.htm 

I hope these make sense and if you or anybody else wish to comment please do so.

Regards
Gyula 



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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #4 on: January 18, 2007, 05:28:44 PM »
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Offline MeggerMan

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #5 on: January 18, 2007, 11:22:34 PM »
Hi Stefan,

Inductance test on a bifilar coil

I have just tried a simple coil wrapped around a small toroidal ceramic core and at 22 turns of 1mm plastic coated wire I get 450uH at 1KHz and 466uH at 120Hz.
The same wire, but wrapped as a bifilar coil around the same core gave me 0.3uH at 1KHz and 0uH at 120Hz.
So yes, provided there is a core it works.

However with a pancake coil wound around in a spiral on a CD the result was as follows.

120uH for the full coil.
27.6uH for one half
28.4uH for the other half.

I then reversed the centre to outer connections (clockwise centre to anti-clockwise outer).
24.1uH for full coil
56.2uH for one half
56.3uH for other half

I need to retest one of my experiments I did the other night now as I realise that the pancake was not connected as a bifilar arrangement now :(

Regards

Rob

Offline MeggerMan

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #6 on: January 19, 2007, 12:28:19 AM »
Hi Gyula,

Quote
One more notice: the ferrite core can be saturated by this huge current and this makes the switch on/off process nonlinear, so the repeatabilty/reconstruction may involve inherent differences in the measured data.  This is nicely shown in Naudin's another test here:

I am wondering if it is the inductance increases as the core is removed and therefore allows a larger back emf. See my tests above.
It would have been useful for JLN to have recorded the coils inductance at each ferrite core position.
As I understand it the inductance will be directly link to the back emf amplitude.
I would have a guess that a bifilar coil with an air core has more inductance than the same coil with a ferrite core or ferrous core.
Could it be the coil windings can each have a very tight localised inductance in air that is disrupted when you introduce a ferrite core, same as if you had lots of single turn coils all coupled in series?
Therefore placing a ferrite core into the coil suddenly cause these single coils to fight against each other and cancel themselves out, hence very little back emf.
 

Regards
Rob

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #6 on: January 19, 2007, 12:28:19 AM »
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Offline pese

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #7 on: January 19, 2007, 09:04:48 AM »
ja , das ist es.  in einer bifilaren spule ist es immer gegeben ,  das beide windungen sich gegenseitig  in einer  transformation entgegenlaufen .  Ein Eisen oder Ferritkern erh?ht diesen Effekt.  Deswegen sinkt der Induktionswert und vermutlich  auch der Wirkungsgrad      GP
yes, that is it. in a bifilaren reel is always given it, which both turns run toward themselves mutually in a transformation. An iron or a ferrite core increases this effect. Therefore the induction value and probably also the efficiency sinks
GP

Offline gyulasun

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #8 on: January 19, 2007, 01:36:25 PM »
Hi Gyula,
...I am wondering if it is the inductance increases as the core is removed and therefore allows a larger back emf. See my tests above.
...As I understand it the inductance will be directly link to the back emf amplitude.
I would have a guess that a bifilar coil with an air core has more inductance than the same coil with a ferrite core or ferrous core.
Could it be the coil windings can each have a very tight localised inductance in air that is disrupted when you introduce a ferrite core, same as if you had lots of single turn coils all coupled in series?
Therefore placing a ferrite core into the coil suddenly cause these single coils to fight against each other and cancel themselves out, hence very little back emf.
 Regards
Rob

Hi Rob,

This morning I made a bifilar solenoid coil similar to that of Naudin's but with less number of turns. My ferrite rod is 9.5mm OD and 170mm long. I covered it with a 55mm long hard paper cylinder of 10mm ID so that it can easily slide on the surface of the rod.
I wound 2 x 23 turns onto the hard paper cylinder from 0.8mm OD enameled copper wire.

I measured the following self inductances:

Any one coil each, no ferrite rod:  L1=L2= 2.3uH
No ferrite rod, L1&L2 in bifilar cancel:       0.42uH
No ferrite rod, L1&L2 in series aiding:        9.1uH

Any one coil each, with ferrite rod, coils slided to any one edge of the rod: L1=L2=35uH
Any one coil each, with ferrite rod, coils slided to the center of the rod:     L1=L2=46uH
With ferrite rod, L1&L2 in bifilar cancel, coils slided to any one edge:                  0.44uH
With ferrite rod, L1&L2 in bifilar cancel, coils slided to the center of the rod:        0.44uH
With ferrite rod, L1&L2 in bifilar aiding, coils slided to any one edge:                   141uH
With ferrite rod, L1&L2 in bifilar aiding, coils slided to the center of the rod:         192uH

So the results show that it does not matter where the bifilar coil pair is placed onto the ferrite rod, its resultant and cancelled inductance (when connected in bifilar cancel mode like in the TEP) remains the same. I am a bit surprised too, that without the ferrite rod the same coils (when connected also in the bifilar cancel mode) show only slightly less self inductance than with the core, i.e. 0.42uH versus the cored 0.44uH.

This may justify the hint on severe core saturation due to the huge peak current and considering Naudin's non-linear time flow tests ( http://jnaudin.free.fr/html/tep61a5.htm )  these show that local, individual turns of wire do influence the core material under them and in the rod center this effect is the greatest (but mainly cancelled) and towards the rod edges and half way out of the coil this saturation effect reduces/diminishes, hence the local self inductances can increase, hence the back emf can also increase. 
Let's not forget that the back emf pulses we see in the scope shots they are there during the transistor switch off time when the coils L1&L2 are floating already, hence already mainly 'recovered' their self inductance. 
Let's also notice that the amplitude of the back emf pulse is very small, under 1V peak when Naudin loaded it with a 100 Ohms resistance (with a diode in series).  For me it may mean that without making further 'tricks' there is no free lunch yet...

Regards
Gyula

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #8 on: January 19, 2007, 01:36:25 PM »
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Offline pese

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #9 on: January 19, 2007, 01:57:49 PM »
Air Coils  , driven with higher frequencies as 1Mhz  will make the same knowledge
that induction will degress or add., the way of the clock (windings) direction.

look for "Variometer" that is an device , that an make this "variable induction"
with moving - to wind- air coil , that build in another -fixed-inside.

(old) RF and Radio-Man?s know that.

PS.  Have look for some pic?s ! Please:
http://www.alg.demon.co.uk/radio/136/pictures/rem_var1.jpg
http://www.historyofpa.co.uk/gfx/cw/crystal/Variometer.jpg
http://w5jgv.com/variometer/variometer.htm  f?r niedrige frequenzen (selbstbau)  low frequencies
 
http://www.vintageradio.info/images/vario-xtal.gif

Pese

@gezin,
think about that , to use fast switching diodes in the bridge rect. (also schottky)
« Last Edit: January 19, 2007, 03:31:18 PM by pese »

Offline hartiberlin

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #10 on: January 19, 2007, 03:08:43 PM »


Hi Rob,

This morning I made a bifilar solenoid coil similar to that of Naudin's but with less number of turns. My ferrite rod is 9.5mm OD and 170mm long. I covered it with a 55mm long hard paper cylinder of 10mm ID so that it can easily slide on the surface of the rod.
I wound 2 x 23 turns onto the hard paper cylinder from 0.8mm OD enameled copper wire.

I measured the following self inductances:

Any one coil each, no ferrite rod:  L1=L2= 2.3uH
No ferrite rod, L1&L2 in bifilar cancel:       0.42uH
No ferrite rod, L1&L2 in series aiding:        9.1uH

Any one coil each, with ferrite rod, coils slided to any one edge of the rod: L1=L2=35uH
Any one coil each, with ferrite rod, coils slided to the center of the rod:     L1=L2=46uH
With ferrite rod, L1&L2 in bifilar cancel, coils slided to any one edge:                  0.44uH
With ferrite rod, L1&L2 in bifilar cancel, coils slided to the center of the rod:        0.44uH
With ferrite rod, L1&L2 in bifilar aiding, coils slided to any one edge:                   141uH
With ferrite rod, L1&L2 in bifilar aiding, coils slided to the center of the rod:         192uH

So the results show that it does not matter where the bifilar coil pair is placed onto the ferrite rod, its resultant and cancelled inductance (when connected in bifilar cancel mode like in the TEP) remains the same. I am a bit surprised too, that without the ferrite rod the same coils (when connected also in the bifilar cancel mode) show only slightly less self inductance than with the core, i.e. 0.42uH versus the cored 0.44uH.


Hi Gyula,
so you think in using your example values,
the Back EMF hill wave of about 0.75 Volts in:
http://jnaudin.free.fr/html/tep61sht.htm
at R3
is caused by the remaining inductance of
0.44 uH in your case ?

Quote

This may justify the hint on severe core saturation due to the huge peak current and considering Naudin's non-linear time flow tests ( http://jnaudin.free.fr/html/tep61a5.htm )  these show that local, individual turns of wire do influence the core material under them and in the rod center this effect is the greatest (but mainly cancelled) and towards the rod edges and half way out of the coil this saturation effect reduces/diminishes, hence the local self inductances can increase, hence the back emf can also increase. 
Let's not forget that the back emf pulses we see in the scope shots they are there during the transistor switch off time when the coils L1&L2 are floating already, hence already mainly 'recovered' their self inductance. 
Let's also notice that the amplitude of the back emf pulse is very small, under 1V peak when Naudin loaded it with a 100 Ohms resistance (with a diode in series).  For me it may mean that without making further 'tricks' there is no free lunch yet...

Regards
Gyula

But why is then one BackEMF smaller at the one side smaller than at the other side,
when the ferrite core is only half way in ?

In your explanation BOTH of them must be bigger, but indeed,
only one is bigger and the other is smaller..??!!

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #10 on: January 19, 2007, 03:08:43 PM »
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Offline gyulasun

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #11 on: January 19, 2007, 03:47:58 PM »
Hi Gyula,
so you think in using your example values, the Back EMF hill wave of about 0.75 Volts in:
http://jnaudin.free.fr/html/tep61sht.htm at R3 is caused by the remaining inductance of
0.44 uH in your case ?


Hi Stefan,

Yes I do.  Remember however that Naudin used 2 x 190 turns  (I used 2 x 23) so his remaining (bifilarly cancelled) inductance was probably higher than 0.44uH I guess.  But it must have been still small, maybe 1-2uH or so.


But why is then one BackEMF smaller at the one side smaller than at the other side,
when the ferrite core is only half way in ?
In your explanation BOTH of them must be bigger, but indeed,
only one is bigger and the other is smaller..??!!

No, I meant the difference in amplitudes when the coils are in the center of the rod both halfs of the rod are nonlinearly saturated more or less in an equal amount and when the coils are at the edges or at half way in only, then the nonlinear saturation shifts towards the edges of the rod and gradually improves (i.e. becomes less and less saturated) towards the center of the rod.  Remember Naudin covered the full length of his rod with the 2 x 190 turns, indicated a 19cm long coil and hence at least as long rod.

When the coils are in the center they cause a equal but opposite saturation or let's say distortion in the core and this mainly is able to cancel out so the resultant amplitude is small, comparing it to the amplitude which comes about when the core is only half way into the coils. In the latter case the the opposite saturation cannot fully be equal because only the half length of the rod is excited so no symmetrical cancellation can take place hence the resultant back emf amplitude is higher with respect to the center case.
That is how I think.  Maybe I am wrong but for the time being I cannot think otherwise.

rgds,  Gyula

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #12 on: January 19, 2007, 03:56:12 PM »
Hi Gyula,

Its good to see you finished off what I started and proved what I suggested to be true.

OK, here's a leap of faith(I'm only guessing here so bare with me):

You pass a current "x" through L1 and L2 with the connections arranged as bifilar.
Then you short out coil L1 so that the current by-passes L1 and goes into L2 only.
I would expect to see the current drop very sharply then rise again as a magnetic field is established in the core.
You will not be able to obtain a back emf of any size until the field is established.
So if you switch off the current just after you short out L1 the back emf will be virtually zero.
However if you wait until the current has risen you will get the full back emf for that coil and current.
So whilst the electron orbits are randomly arranged in the core there is no stored energy.
Once they re-align then you can turn off the current, the driver for the field collapses and the orbits are then free to return to a random state and induce a back EMF.

Hi Stefan,
Quote
In your explanation BOTH of them must be bigger, but indeed,
only one is bigger and the other is smaller..??!!

I think you may be mistaken, the middle position of the ferrite core produces the smallest back emf and when the core is either of the two outer position the back emf is greater.

Regards
Rob

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #13 on: January 19, 2007, 04:06:14 PM »
Hi rob,

So you mean L1&L2 is in bifilar cancel connection and you short out one of them while current flows in both of them, right?

Well, current would only change significantly if your switch inner resistance is much smaller then either L1 or L2 copper resistance, ok?
Then next problem I see is that when you short one half of bifilarly connected coil pair then you kill the self inductance of both!  Would not be this matter in your idea?

(Must leave now, will be back tonight )

Gyula

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Re: Charging a coil with less energy and get huge BackEMF energy pulses
« Reply #14 on: January 19, 2007, 04:49:22 PM »
Hi Gyula,
Quote
So you mean L1&L2 is in bifilar cancel connection and you short out one of them while current flows in both of them, right?
Yes

Quote
Well, current would only change significantly if your switch inner resistance is much smaller then either L1 or L2 copper resistance, ok?
I was thinking that this is for a pulsed circuit say and the frequency is that for a near perfect sine wave, albeit only half a wave.

If you have no inductance when current is flowing through L1 and L2 and then you short L1, the field in the core will be non-existent and the current will rise to build up the field, I think, I may be wrong of course.
Only an experiment will prove this.
Rob

 

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