# The Naked Scientists Forum

### Author Topic: Why wouldn't this work? (perpetuum motion)  (Read 18757 times)

#### AlmostHuman

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##### Why wouldn't this work? (perpetuum motion)
« Reply #25 on: 22/06/2011 19:03:42 »
@Geezer

Potential energy for charged particle/object in electric field is Wp=Q*E (Where Q is charge of particle/object, and E is module of vector of electric field). Now, if we do the math... we would see that this electric field isn't really as conservative as we thought it is. What do I mean by that? Let's try shooting our particle from point A between capacitor plates (like shown before) and see where particle goes and take one point in that trajectory and mark it as B. Now, if we slightly move our trajectory (try to hit B in straight line) we would see that we need to do some work that was done by capacitor to actually hit the B point again. Which shouldn't be "allowed" if this was truly conservative field. This was just to question that path independent statement.

@JP
Are you saying that Moon/Earth are losing THEIR potential energy, since they are doing all that work?
The work that was done by capacitor field wouldn't be done if we didn't take a path between the plates, therefore, the work
was done by capacitors energy, not by particle's potential energy.

PS
All I want to know is where this energy comes from? And I think that our discussion is progressing quite nicely, because I think we are getting somewhere :).
I wanted this discussion to come to the point where we would try to combine two independent fields to work together, and create useful work. And since I will be occupied for next few days (I don't know if I will be able to participate here next three to five days) I'm jumping maybe too far ahead. (We didn't clear out that energy thing)

#### Geezer

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##### Why wouldn't this work? (perpetuum motion)
« Reply #26 on: 22/06/2011 20:40:46 »
Quote
Geezer said: "In running round ANY closed loop in a static electric field the field

No he didn't. That was some other geezer.

#### Geezer

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##### Why wouldn't this work? (perpetuum motion)
« Reply #27 on: 23/06/2011 00:54:18 »

All I want to know is where this energy comes from?

It doesn't. The only energy present is the energy that was initially stored in capacitor, and a lot of that was dissipated in the loop when it was introduced between the plates.

I suggest you construct a model, test it, and publish the results before we move this topic to "New Theories"

#### Soul Surfer

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##### Why wouldn't this work? (perpetuum motion)
« Reply #28 on: 23/06/2011 08:44:50 »
A charged particle moving between one plate and the other of a charged capacitor is just the same as the electrical breakdown of that capacitor,  that is the capacitor is slightly discharged.  In the case of the non contacting loop of conductor through the capacitor as I have stated above the charge in the conducting wire just rearranges itself to match the potential of the locality clearly if it is a loop the large voltage drops quickly between the plates and builds slowly round the loop.  The energy of this rearrangement depends on the self capacitance of the wire and the rate at which it does the rearranging when the capacitor is charged depends on the self inductance of the wire loop.

I state again the total voltage drop around any closed loop in a static electric field is always zero.

#### JP

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##### Why wouldn't this work? (perpetuum motion)
« Reply #29 on: 23/06/2011 20:00:38 »
I'm  not quite sure moving a charge is the same as discharging the capacitor.  In the figure he showed above, the charge is introduced separately from the capacitor itself and doesn't touch the plates.  The charge on the plates shouldn't change and once the independent charge has been removed, the energy stored in the field between the plates should be the same as it was initially.

If the plates were slightly discharged, the charge on the plates would change and the field between them would be reduced, reducing the total energy stored between them.

#### JP

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##### Why wouldn't this work? (perpetuum motion)
« Reply #30 on: 23/06/2011 20:19:31 »
Maybe we can simplify the question this way:

You have a parallel plate capacitor charged up and disconnected so that the charges on the plates can't escape.  You then fire a charge between the plates on a path that's initially parallel to them.  It has high enough velocity that it makes it through the plates even though it's deflected by them.

Coming out, it would seem that it's velocity should be higher than going in, so it gained kinetic energy.  In addition, a charge was accelerated so it should radiate away some energy.  If the charge on the plates remained the same, the field between them should be the same when the particle has gone on it's merry way.  Where did this extra kinetic and radiative energy come from, then?

#### Geezer

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##### Why wouldn't this work? (perpetuum motion)
« Reply #31 on: 24/06/2011 04:49:21 »
It's really just a capacitor with a conductor looping around one of the plates. When a potential difference is applied across the plates they will charge with opposite potentials. While the plates are charging and the electric field strength is increasing, a small transient current will be induced in the loop.

When the plates are fully charged there is no potential difference between the ends of the diagonal wire because the field strength at both ends of the wire is exactly the same (although that may not be obvious).

It may help to consider what happens if the loop is connected to ground. You then have the equivalent of three capacitors. Two go between ground and each plate connection, and a third goes between the plate connections.

Here is the equivalent circuit;

which, of course, is equivalent to this;

or this;

« Last Edit: 24/06/2011 06:45:24 by Geezer »

#### AlmostHuman

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##### Why wouldn't this work? (perpetuum motion)
« Reply #32 on: 26/06/2011 21:57:46 »
Maybe we can simplify the question this way:

You have a parallel plate capacitor charged up and disconnected so that the charges on the plates can't escape.  You then fire a charge between the plates on a path that's initially parallel to them.  It has high enough velocity that it makes it through the plates even though it's deflected by them.

Coming out, it would seem that it's velocity should be higher than going in, so it gained kinetic energy.  In addition, a charge was accelerated so it should radiate away some energy.  If the charge on the plates remained the same, the field between them should be the same when the particle has gone on it's merry way.  Where did this extra kinetic and radiative energy come from, then?

Thank you JP! That was the source of my confusion... and it still is :). I need someone really good with this stuff to explain to me how the capacitor lose it's energy by that process.

PS If this particle had significant mass, and if it was deflected upwards (gravity perspective) we could get some work done, right?

#### Geezer

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##### Why wouldn't this work? (perpetuum motion)
« Reply #33 on: 27/06/2011 09:55:18 »

I need someone really good with this stuff to explain to me how the capacitor lose it's energy by that process.

It can't. The capacitor didn't lose any energy that wasn't there in the first place. The extra loop of wire forms part of the capacitor and, by a remarkable coincidence, it actually determines the total capacitance of the capacitor, including the loop of wire.

When you charge the capacitor with that particular capacitance to any potential, assuming the dielectric of the capacitor has no losses, the energy that went into charging that particular capacitor is still in the capacitor.

As the title of this topic implies some sort of "free energy", it has been moved to new theories. However, a new topic could be started to help understand the physical mechanisms involved within this type of capacitor.

#### JP

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##### Why wouldn't this work? (perpetuum motion)
« Reply #34 on: 27/06/2011 14:30:50 »

I need someone really good with this stuff to explain to me how the capacitor lose it's energy by that process.

It can't. The capacitor didn't lose any energy that wasn't there in the first place. The extra loop of wire forms part of the capacitor and, by a remarkable coincidence, it actually determines the total capacitance of the capacitor, including the loop of wire.

Yes, that's true for the wire.  But what's the explanation if you send a single charged particle through the capacitor as described above?  I can hand-wave a bit and talk about fields that aren't between the two plates and conservation of energy, but I feel like there should be a much more elegant answer that I'm missing...

#### Geezer

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##### Why wouldn't this work? (perpetuum motion)
« Reply #35 on: 27/06/2011 18:58:40 »

I need someone really good with this stuff to explain to me how the capacitor lose it's energy by that process.

It can't. The capacitor didn't lose any energy that wasn't there in the first place. The extra loop of wire forms part of the capacitor and, by a remarkable coincidence, it actually determines the total capacitance of the capacitor, including the loop of wire.

Yes, that's true for the wire.  But what's the explanation if you send a single charged particle through the capacitor as described above?  I can hand-wave a bit and talk about fields that aren't between the two plates and conservation of energy, but I feel like there should be a much more elegant answer that I'm missing...

If I understand the setup properly, I'm not sure a charged particle would be deflected at all if the plates of the capacitor were disconnected from everything. In that situation I think the net charge on the particle is zero with respect to the capacitor.

It would be like trying to deflect electrons from the gun in a cathode ray tube with deflection plates that were not referenced to the electrons' potential. The electrons will ignore any potential on the plates and fly straight ahead.

#### JP

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##### Why wouldn't this work? (perpetuum motion)
« Reply #36 on: 27/06/2011 20:12:28 »

I need someone really good with this stuff to explain to me how the capacitor lose it's energy by that process.

It can't. The capacitor didn't lose any energy that wasn't there in the first place. The extra loop of wire forms part of the capacitor and, by a remarkable coincidence, it actually determines the total capacitance of the capacitor, including the loop of wire.

Yes, that's true for the wire.  But what's the explanation if you send a single charged particle through the capacitor as described above?  I can hand-wave a bit and talk about fields that aren't between the two plates and conservation of energy, but I feel like there should be a much more elegant answer that I'm missing...

If I understand the setup properly, I'm not sure a charged particle would be deflected at all if the plates of the capacitor were disconnected from everything. In that situation I think the net charge on the particle is zero with respect to the capacitor.

It would be like trying to deflect electrons from the gun in a cathode ray tube with deflection plates that were not referenced to the electrons' potential. The electrons will ignore any potential on the plates and fly straight ahead.

Hmm... I suspect we don't understand the setup in the same way, then.  The way I read it is that you've basically charged the two plates up then disconnected them while still charged.  With nowhere else to go, the charge remains on the plates, so you have two parallel plates with equal but opposite charges.  If you introduce a charged particle between them, it will feel a force.  If it's an electron, the force will point from the negative to the positive plate.

My question is whether there's an easy way to explain the conservation of energy in the case where you fire a particle through the two plate system so that it escapes out the other side.  Simplistically you'd think that the particle would pick up extra velocity as a result of this force it feels between the plates, so it has more energy coming out than going in.  The charge on the plates doesn't change.  Therefore, how is energy conserved?  The coil is a poisson rouge so far as this question goes.

It is, of course, conserved, and I suspect it has to do with taking into account motion as well as magnetic fields and also realizing the fields aren't confined entirely to the region between the plates.  It feels like there should be a more straightforward explanation, but it's escaping me...

#### Geezer

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##### Why wouldn't this work? (perpetuum motion)
« Reply #37 on: 27/06/2011 20:35:10 »
Hmm... I suspect we don't understand the setup in the same way, then.  The way I read it is that you've basically charged the two plates up then disconnected them while still charged.  With nowhere else to go, the charge remains on the plates, so you have two parallel plates with equal but opposite charges.  If you introduce a charged particle between them, it will feel a force.

I don't think it will feel any force at all

As the plates are not referenced to anything, the plates will reference themselves to the charged particle. The potential on the particle (from the plates) will be zero.

In reality there would be some capacitance between the plates and ground, so they would have some common reference, but as described, there is no common reference. The plates are truly "floating".

Think of it this way - if you measure the voltage between either of the plates and ground, it will always be zero. The electron does the same thing.

#### JP

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##### Why wouldn't this work? (perpetuum motion)
« Reply #38 on: 27/06/2011 20:51:15 »
The plates will be at different potentials with respect to each other.  Ground is irrelevant here.  But you don't even need potential at all.  Two parallel plates of opposite charges create a field in the gap.  A charged particle in a field experiences a force.

Of course, the plates need to be mechanically held in place (but not electrically grounded) and the charge free to move, but I believe that was stated in the original description of the system.

#### Geezer

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##### Why wouldn't this work? (perpetuum motion)
« Reply #39 on: 27/06/2011 21:31:21 »
B-b-b-but, if a particle carries charge, it has to be charged with respect to something. So, you can only measure the charge on a particle with respect to some reference. There's no "absolute" charge - at least I don't think there is.

In this case, as the plates and the particle share no common reference, it's the particle that's the "immovable" object. The capacitor adjusts it's potentials to be relative to the particle rather than the other way around.

If the particle was charged to a million volts (relative to some reference), when the particle shot between the plates, the plates would be elevated to a million volts relative to the particle's reference, plus and minus their relative potentials.

#### JP

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##### Why wouldn't this work? (perpetuum motion)
« Reply #40 on: 27/06/2011 21:43:54 »
B-b-b-but, if a particle carries charge, it has to be charged with respect to something. So, you can only measure the charge on a particle with respect to some reference. There's no "absolute" charge - at least I don't think there is.

There is an absolute charge, though.  Charge isn't like voltage, where everything is defined with respect to a reference.  Charge is a fundamental physical quantity that comes in discrete units.  If I have a ball with 1 Coulomb of charge and one of 2 Coulombs of charge repelling each other, the resulting physics is quite different than if I had 0 Coulombs and 1 Coulomb or +1/2 Coulomb and -1/2 Coulomb, so you can't just pick an arbitrary reference.  (Zero charge physically means that there are as many negative as positive charges in something and negative and positive charges are fundamental properties of subatomic particles.)

#### Geezer

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##### Why wouldn't this work? (perpetuum motion)
« Reply #41 on: 27/06/2011 22:01:25 »
Ah yes, but charge and potential are not the same things, and it's the potential that produces the deflection. In this case, there is no potential between the particle and the plates, so there is no deflection.

#### Geezer

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##### Why wouldn't this work? (perpetuum motion)
« Reply #42 on: 27/06/2011 22:27:27 »
Ah yes, but charge and potential are not the same things, and it's the potential that produces the deflection. In this case, there is no potential between the particle and the plates, so there is no deflection.

I should be a bit more precise. The point where the particle is in the capacitor's field will have a potential equal to the potential of the particle when it was charged relative to some reference. The plates will be at potentials relative to that reference because that's the easiest thing for them to do, and there is nothing to prevent them assuming those potentials.

If the particle is "half way" between the plates, the plates will have equal positive and negative potentials relative to the potential of the particle. If the particle is closer to the positive plate, the negative plate will have a greater absolute potential (more negative) and so on. Whatever happens, any forces experienced by the particle because of the electric field will be equal and opposite because the potentials of the plates adjust to the potential of the particle.

(Phew! - Wipes sweat from forehead.)

#### burning

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##### Why wouldn't this work? (perpetuum motion)
« Reply #43 on: 27/06/2011 22:28:57 »
Ah yes, but charge and potential are not the same things, and it's the potential that produces the deflection. In this case, there is no potential between the particle and the plates, so there is no deflection.

Um, no.  Stop thinking about potential for a moment and think about charges and electric fields.  As JP pointed out, charge is absolute, and we really can charge a capacitor so that one plate has an excess of positive charge and the other plate has an equal magnitude excess of negative charge.  These physically separated, non-zero charge distributions will produce an electric field.  This electric field will be strongest between the plates (although it will be present elsewhere).  If we fire a charged particle in between the plates it will experience an electric force (it experiences it before it gets between the plates, but the force is strongest between the plates).  That force will produce a change in kinetic energy of the particle.  Now, since the field is non-zero outside, the net work done on the particle very much depends on where it started and where it ends up, but it will definitely be non-zero for a large set of finite trajectories.

OK, now what about the electrostatic potential?  It is true that absolute electrostatic potential has no meaning.  It is arbitrary where we define 0 potential.  However, electrostatic potential difference is another matter.  The potential difference between two points is equivalent to the work done by the field per unit charge on a particle travelling between those points.  To assert that this difference is nonexistent is tantamount to saying that there is no electric field.

#### burning

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##### Why wouldn't this work? (perpetuum motion)
« Reply #44 on: 27/06/2011 22:32:45 »
If the particle is "half way" between the plates, the plates will have equal positive and negative potentials relative to the potential of the particle. If the particle is closer to the positive plate, the negative plate will have a greater absolute potential (more negative) and so on. Whatever happens, any forces experienced by the particle because of the electric field will be equal and opposite because the potentials of the plates adjust to the potential of the particle.

(Phew! - Wipes sweat from forehead.)

Except the plates have opposite charges.  Let's say we have the negatively charged plate on the left and the positively charged plate on the right.  Now let's fire a negatively charged particle between the plates.  It is repelled by the negatively charged plate, so it is pushed to the right.  It is attracted by the positively charged plate so it is pulled to the right.  These forces do not cancel, they work together.

#### Geezer

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##### Why wouldn't this work? (perpetuum motion)
« Reply #45 on: 27/06/2011 22:57:57 »
I don't think so Burning.

You seem to be assuming the plates have some constant potential relative to the particle. If they did, the particle would certainly be deflected. However, while the relative potentials of the plates remains constant, their mean potential can have any arbitrary value at any instant, and the particle can change their mean potential because there is absolutely nothing to prevent it from doing so.

If the particle did deflect in this situation, it would be analogous to operating a lever without a fulcrum

#### burning

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##### Why wouldn't this work? (perpetuum motion)
« Reply #46 on: 28/06/2011 02:20:26 »
I don't think so Burning.

You seem to be assuming the plates have some constant potential relative to the particle.

No. I am stating that if the capacitor has been charged and then disconnected from the circuit, as specified by the OP, then there is a constant potential difference between the two plates.  If we fire a charged particle through the electric field it will not maintain a constant potential difference with either plate.

Quote
If they did, the particle would certainly be deflected.  However, while the relative potentials of the plates remains constant, their mean potential can have any arbitrary value at any instant, and the particle can change their mean potential because there is absolutely nothing to prevent it from doing so.

If the particle did deflect in this situation, it would be analogous to operating a lever without a fulcrum

Sorry, I can't understand what you mean by the rest of this.  I ask again that we step away from the electrostatic potential and look at the charges and the electric field.  Which if any of the following assertions do you consider to be false and why?

1.) It is possible to charge a capacitor, by which we mean that we create a net negative charge on one plate and a net positive charge of equal magnitude on the other plate.

2.) An object with a net non-zero charge will produce an electric field

3.) In the case of the capacitor, the electric field is strongest between the plates.  This is because the direction of the field from the positive plate matches that of the field from the negative plate in this region.

4.) If a charged particle passes through a region with an electric field, it will experience an electric force.

5.) In the case of a charged particle between the plates of a capacitor, it will be attracted toward the plate with the opposite sign and be repelled by the plate with the same sign.

#### JP

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##### Why wouldn't this work? (perpetuum motion)
« Reply #47 on: 28/06/2011 04:37:39 »
Geezer, I think burning is absolutely right and if you can point out which of his points is wrong, maybe we can understand why we disagree.  In the meantime, I also have a specific disagreement with one thing you said:

Ah yes, but charge and potential are not the same things, and it's the potential that produces the deflection. In this case, there is no potential between the particle and the plates, so there is no deflection.

Force is what produces the deflection.  Potential is not force.  Potential is a useful computational tool for getting the electric field and force may be computed from the electric field (in the statics case).  It's very close to potential energy, since it's the integral of the field over distance.  That's why you can arbitrarily re-zero potential energy by picking a reference point.  Just like potential energy, it's differences that matter, not magnitude.  You can't do the same with field or with electric forces.

http://en.wikipedia.org/wiki/Electric_potential#In_electrostatics
« Last Edit: 28/06/2011 05:51:46 by JP »

#### Geezer

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##### Why wouldn't this work? (perpetuum motion)
« Reply #48 on: 28/06/2011 07:36:01 »
Geezer, I think burning is absolutely right and if you can point out which of his points is wrong, maybe we can understand why we disagree.  In the meantime, I also have a specific disagreement with one thing you said:

Ah yes, but charge and potential are not the same things, and it's the potential that produces the deflection. In this case, there is no potential between the particle and the plates, so there is no deflection.

Force is what produces the deflection.  Potential is not force.  Potential is a useful computational tool for getting the electric field and force may be computed from the electric field (in the statics case).  It's very close to potential energy, since it's the integral of the field over distance.  That's why you can arbitrarily re-zero potential energy by picking a reference point.  Just like potential energy, it's differences that matter, not magnitude.  You can't do the same with field or with electric forces.

http://en.wikipedia.org/wiki/Electric_potential#In_electrostatics

Yes - I agree. That's why I corrected my post.

It will take me some time to respond to Burnings points. Stay tuned

#### Geezer

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##### Why wouldn't this work? (perpetuum motion)
« Reply #49 on: 28/06/2011 21:28:19 »
OK - Here we go again.

I don't think so Burning.

You seem to be assuming the plates have some constant potential relative to the particle.

Quote
No. I am stating that if the capacitor has been charged and then disconnected from the circuit, as specified by the OP, then there is a constant potential difference between the two plates.

Yes - I fully agree with you there.

Quote
If we fire a charged particle through the electric field it will not maintain a constant potential difference with either plate.

That's where it gets tricky. I'll come back to that.

Quote
If they did, the particle would certainly be deflected.  However, while the relative potentials of the plates remains constant, their mean potential can have any arbitrary value at any instant, and the particle can change their mean potential because there is absolutely nothing to prevent it from doing so.

If the particle did deflect in this situation, it would be analogous to operating a lever without a fulcrum

Sorry, I can't understand what you mean by the rest of this.  I ask again that we step away from the electrostatic potential and look at the charges and the electric field.  Which if any of the following assertions do you consider to be false and why?

Quote
1.) It is possible to charge a capacitor, by which we mean that we create a net negative charge on one plate and a net positive charge of equal magnitude on the other plate.

Completely agree.

Quote
2.) An object with a net non-zero charge will produce an electric field

If it's charged, yes, and in this case it certainly is.

Quote
3.) In the case of the capacitor, the electric field is strongest between the plates.  This is because the direction of the field from the positive plate matches that of the field from the negative plate in this region.

Completely agree.

Quote
4.) If a charged particle passes through a region with an electric field, it will experience an electric force.

I'd say it slightly differently. I'd say that it will definitely interact with the electric field and behave according to a very specific set of rules.

Quote
5.) In the case of a charged particle between the plates of a capacitor, it will be attracted toward the plate with the opposite sign and be repelled by the plate with the same sign.

Disagree. That does not seem to take into account the potential at the point in the field and the potential of the particle. It's entirely possible for the particle to travel through the capacitor without experiencing any deflection.

I've thought of several ways to explain what I think is happening, but this might be the most straightforward and hopefully it will either make sense, or fall apart quite rapidly.

As Burning points out above, the capacitor as a whole is "neutral" - it has no net positive or negative charge. Consider it to be, as the diagram implies, totally isolated from everything after it has been charged up.

Now, along comes a charged particle with a potential of +x volts that has travelled from a great distance. The charged particle enters the electric field of the capacitor and, if I have it right, the potential of the electric field at that point will now be +x volts (I think that's the most crucial part of my argument.)

Also, I think the capacitor as a whole is no longer "neutral". It now has a charge equal to the charge of the particle. The potential difference between the plates does not change, but the potentials on the plates are now relative to the potential of the particle.

Let's see if we can blow up that description before we go any further.
« Last Edit: 28/06/2011 21:39:12 by Geezer »

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##### Why wouldn't this work? (perpetuum motion)
« Reply #49 on: 28/06/2011 21:28:19 »