# Continuous Frictioned Motion Machine

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### #41 christopherkirkreves

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Posted 8 February 2011 - 07:28 PM

John Cuthber : " " (The unknowns are becoming less.)" The only unknown is why do you still believe in this. "

However, I appreciate even more your analysis of my idea being the same idea as Bishop's, even though I disagree.

John Cuthber: " Since you don't seem to wish to accept reality I guess I should point out one reason why your system (or the 17th C Bishop's) won't work. If the magnet at the top left is strong enough to pull the moving thing ( ferrofluid or steel shot) across from the top righ of the diagram to the top left or up the slope, then it will be too strong to let that moving thing go somewhere else- like the bottom of the diagram, or down through the hole in the slope. It's different in detail; but the reason it won't work is exactly the same and, if you had thought about the Bishop's "mechanism" and why it fails you would have realised why you system would also fail. "

It is possible to set up Bishop's design (  I've tried it , and learned from experiment  ) and get the metal ball to roll up and the ramp  and to also get the ball to fall through the hole. By pulling the ball up the ramp, the force on the ball from gravity is much less than the force on the ball from gravity when exposed to the hole and a free fall. You can move the ball towards the magnet and then away from the magnet. However, while you can get the ball to fall, you cannot get the ball all the way back to the base of the ramp. As the ball falls, it falls more slowly than it otherwise would because of the magnet pulling up on it. And after falling, as the ball starts to roll back towards the base of the ramp, it's still in the magnetic field, and so there is a backwards pull on it from the magnet. If you think about it, this idea is simply an attempt to use a magnetic field to overcome a gravitational field, and then an attempt to use that same gravitational field to overcome the same magnetic field. It would work  if there was no friction (  like a hypothetical frictionless pendulum swinging back and forth forever).

The idea I have proposed here, to my mind, is something quite different.

"If the magnet at the top left is strong enough to pull the moving thing ( ferrofluid or steel shot) across from the top righ of the diagram to the top left or up the slope, then it will be too strong to let that moving thing go somewhere else- like the bottom of the diagram, or down through the hole in the slope."

After the ferrofluid drips across the gap, it is on the face of the second magnet, and at the weakest (thinnest) part of the second magnet. It makes logical sense that the ferrofluid will then move to the greatest (thickest) part of the second magnet, and I have demonstrated that this does actually happen.

I think the difference between this idea and Bishop's idea in somewhere in the area of in this idea I'm not simply using the magnets to pull the ferrofluid towards it and away from another force. Rather, here I make use of the ferrofluids natural tendency to spike along the lines of flux. (When a ferrofluid spikes out along the lines of flux and away from a magnet, it moves out from the greater part of the magnetic field to a lesser part.) And, in this design, when the ferrofluid spikes out at the end of the capillary, it is spiking away from the strongest (thickest) part of the first magnet towards the weakest (thinnest) part of the second magnet. However, I suspect that after moving through the capillary and spiking out, since it is so much closer to the weakest part of the second magnet than the strongest part of the first magnet, that at this point it is spiking towards, into, a greater magnetic field. There is more pull on the fluid from the weakest part of the second magnet, than there is from the strongest part of the first magnet, due to the fluid being so much closer to the second magnet.

And all of this happens in the presence of friction.

Do you still believe that these two machines are simply two versions of the same thing? If so  then  I guess we'll just have to disagree.

Thank you.

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### #42 John Cuthber

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Posted 8 February 2011 - 07:36 PM

Fundamentally, all perpetual motion machines are the same thing- a doorstop.
The differences are, as I say, in the detail and you just haven't understood it.
The ferrofluid will be attracted to the strongest part of the magnet and, because that's the place it's most attracted to, it will stay there.
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### #43 Ethereally Luminous

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Posted 9 February 2011 - 01:15 AM

It does seem like no matter how many ideas for perpetual mechanisms there is always some inhibiting factor that balances the mechanism to a static phase. Don't give up though if you created a perpetual motion device you'd go down in history as the founder of The Law of Perpetual Motion .
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### #44 christopherkirkreves

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Posted 12 February 2011 - 11:33 PM

Ethereally Luminous:

"It does seem like no matter how many ideas for perpetual mechanisms there is always some inhibiting factor that balances the mechanism to a static phase."

I agree. And I can understand why the folks in this forum simply don't believe me. However, if this machine will reach a state of balance, then there should be a simple clear way to state why, such as:

1. The ferrofluid will not move to the strongest part of the magnetic field.

2. The ferrofluid will not move into the capillary and away from the strongest part of the magnetic field.

3. The ferrofluid will not spike out beyond the end of the capillary.

4. The ferrofluid will not break off from the end of the capillary and drip onto the weakest part of the second magnet.

or

5. The ferrofluid necessarily becomes corrupted after moving through the first capillary in a way that will not allow it to move into a second capillary.

I've posted another video on You Tube. It's titled "14th Machine. A one sided attempt."

In this experiment, I start by putting the ferrofluid on only one of the two wedge magnets.

The fluid is pulled to the greatest part of the first magnet. The ferrofluid moves into the first capillary. The ferrofluid reaches the end of the capillary and spikes. The ferrofluid breaks off from the end of the capillary. The ferrofluid drips onto the weakest part of the second magnet. The ferrofluid is pulled to the strongest part of the second magnet. And, the ferrofluid enters the second capillary.

Unfortunately, after four days of slowly dripping, the first capillary stops dripping due to evaporation, so there is no more fluid available to enter the second capillary, and the fluid only enters the second capillary to about a half a centimeter.

However, this experiment does show that the ferrofluid does not become corrupted when moving through the first capillary in a way that keeps it from entering the second.

"Don't give up though "

Thank you. I won't.

John Cuthber:

"The ferrofluid will be attracted to the strongest part of the magnet and, because that's the place it's most attracted to, it will stay there."

After coming up with this design, but before I built it, I suspected that this might be true. I suspected that perhaps after reaching the strongest part of the magnet, that the magnetic liquid would not move into the capillary and away from the strongest part of the magnet. But it does. And this can be seen happening in the videos on You Tube. (It's is especially clear in the latest one: "14th Machine. A one sided attempt.")

If you're interested, you can see the ferrofluid move away from the strongest part of the magnet.

Thank you.

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### #45 md65536

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Posted 13 February 2011 - 02:55 AM

"Don't give up though "

Thank you. I won't.

I disagree with that advice. I'd recommend researching a bit more so that you understand why these machines don't work, instead of trying multiple times to build them.
I think that it is very unlikely that you can break a law of science without first understanding it enough to know where (and only IF) it can be broken.

You requested and listed possible explanations for why your machine doesn't work, but you've skipped the suggestions given to you.

Here is your flaw: You are starting your machine in one state (capillaries empty; relatively high amount of potential energy) and then letting it run to another state (capillaries saturated) at which point, for some reason you expect it to keep running.

Once you get to a stable state, there are not really any "new states" for the machine to get to. You are claiming that the machine is in one state, the liquid drips, the capillaries suck up more liquid, and you are back at a state very similar to the previous state.

So answer this: What form of energy is used to return the machine to a previous state?
If you can answer that realistically, then you've identified an energy input to your machine.

It's nothing special that the machine would go from its start state to a stable state, even if that involves the liquid dripping and moving across the magnet, even cycling for days or whatever else it might do. However I don't think you've shown that your machine goes from a stable state, through a cycle, and returns to a stable state. Without energy input this is impossible. Yet you're ignoring this, and looking for other excuses for why the machine stops working.
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### #46 John Cuthber

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Posted 13 February 2011 - 11:36 AM

Fundamentally there is somewhere in the system that is the "best" place for the fluid to be.
It might be at the pole of the magnet, or it might be in the tube. But it can't be both.
If it's in the tubes then it will go there and stop, If it's at the pole of the magnet it will get to the pole and stop.

If you really think the problem is due to evaporation, put the thing in a plastic bag with a wet sponge.

When it still stops, you will have to dream up another excuse for it.
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### #47 christopherkirkreves

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Posted 16 February 2011 - 06:34 PM

excuses, excuses, excuses

md65536: "Yet you're ignoring this, and looking for other excuses for why the machine stops working."

John Cuthber: "When it still stops, you will have to dream up another excuse for it."

As these machines continue to drip day after day, the spikes become longer and longer before breaking off. And, at some point, the spikes become long enough to bridge all the way across the gap. At this point the systems come to a stop. And, if you poke at the bridged ferrofluid, it's brittle, and snaps off. It no longer acts like a liquid. (This is also true of the other spikes of ferrofluid along the magnets themselves. They become brittle and non fluid like after a few days.)

My guess is that the fluid part of the ferrofluid evaporates, leaving a greater concentration of suspended solids, and so they act more like a collection of solids than a fluid.

There is another possibility. Perhaps as the ferrofluid moves through the capillaries they become corrupted in a way that makes the fluid brittle. If that is so, then there must be a specific reason why a ferrofluid moving through a capillary becomes less fluid like and more like a collection of solids.

I'm open to the discussion.

John Cuthber: "If you really think the problem is due to evaporation, put the thing in a plastic bag with a wet sponge."

Yes. You're right. It's time for a 15th machine.

md65536: "You requested and listed possible explanations for why your machine doesn't work, but you've skipped the suggestions given to you."

I hope not.

md65536: "You are claiming that the machine is in one state, the liquid drips, the capillaries suck up more liquid, and you are back at a state very similar to the previous state."

Yes, and I think I've confused the issue by talking about the part where I put the ferrofluid into the system before the capillaries are saturated.

Time One. Both capillaries are saturated with ferrofluid. And, there is ferrofluid outside the capillaries, but in contact with the capillaries, at the base of both capillaries.

Time Two. The ferrofluid spikes out of the capillaries towards the weakest parts of the other magnets.

(The question is: what happens then? It was suggested in this forum that this will create an empty space within the capillaries. If this is true, then my design is a failed design. If the capillaries fill up, drain, but do not fill up again, then my machine does not work. Cool. But I doubt that is what is happening. I suspect that as the ferrofluid is pulled out beyond the end of the capillary, that, due to cohesion, more fluid is pulled into the capillaries from the surrounding fluid at the base of the capillaries (fluid outside of the capillaries, but in contact with the fluid in the capillaries). Just as a tree does not take in fluid once, then loses it through the leaves, and thus becomes empty of water. A tree will take in more water from the surrounding soil due to osmosis and cohesion.)

Time Three. The ferrofluid breaks off and drips.

Time Four. The ferrofluid moves to the strongest part of the magnet and joins the rest of the ferrofluid there (outside the next capillary, but in contact with it).

md65536: "So answer this: What form of energy is used to return the machine to a previous state? If you can answer that realistically, then you've identified an energy input to your machine."

After this system is set up, I do not see any additional energy inputs into it.

md65536: "However I don't think you've shown that your machine goes from a stable state, through a cycle, and returns to a stable state."

Stable state. The capillaries are saturated with ferrofluid and surrounded by ferrofluid at their bases.

Cycle. The ferrofluid spikes out of the capillaries. The ferrofluid breaks off and drips. The ferrofluid moves to the strongest part of the next magnet and joins the ferrofluid there at the base of the other capillary.

Stable state. The capillaries are saturated with ferrofluid and surrounded by ferrofluid at their bases.

(That is, of course, assuming that I'm right in that as the fluid is pulled out of the capillary that more fluid will move into the capillary (thus keeping it saturated) due to cohesion. I'm open to the discussion. Why would more fluid not move into the capillaries at their bases, as fluid spikes out at the tops?)

John Cuthber: "If it's in the tubes then it will go there and stop, If it's at the pole of the magnet it will get to the pole and stop."

The fluid does move into the tube and away from the pole of the magnet, and the fluid does also move out of the tube and back onto the pole of the other magnet. It happens.

Thank you.
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### #48 michel123456

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Posted 16 February 2011 - 06:46 PM

Maybe this?

http://en.wikipedia....terial_behavior

"As mentioned above, smart fluids are such that they have a low viscosity in the absence of an applied magnetic field, but become quasi-solid with the application of such a field. In the case of MR fluids (and ER), the fluid actually assumes properties comparable to a solid when in the activated ("on") state, up until a point of yield (the shear stress above which shearing occurs). This yield stress (commonly referred to as apparent yield stress) is dependent on the magnetic field applied to the fluid, but will reach a maximum point after which increases in magnetic flux density have no further effect, as the fluid is then magnetically saturated."
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Michel what have you done?

### #49 md65536

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Posted 16 February 2011 - 10:53 PM

My guess is that the fluid part of the ferrofluid evaporates, leaving a greater concentration of suspended solids, and so they act more like a collection of solids than a fluid.

Evaporation typically involves an energy input.

Stable state. The capillaries are saturated with ferrofluid and surrounded by ferrofluid at their bases.

Cycle. The ferrofluid spikes out of the capillaries. The ferrofluid breaks off and drips. The ferrofluid moves to the strongest part of the next magnet and joins the ferrofluid there at the base of the other capillary.

Stable state. The capillaries are saturated with ferrofluid and surrounded by ferrofluid at their bases.

(That is, of course, assuming that I'm right in that as the fluid is pulled out of the capillary that more fluid will move into the capillary (thus keeping it saturated) due to cohesion. I'm open to the discussion. Why would more fluid not move into the capillaries at their bases, as fluid spikes out at the tops?)

Exactly! This is the key to understanding your device! How can any system perpetually cycle through the same states without requiring any new energy, and while losing (a tiny amount of) energy due to friction? Conservation of energy is what you need to research. If you can extract a certain amount of energy going from state A to B, it will take at least that much energy to go from state B (through any states C etc) back to state A.

- A system can cycle for a very long time if it is very efficient. A pendulum won't swing to the bottom and then immediately stop, even though it has the lowest potential energy there, because potential energy is constantly being converted to kinetic energy and vice versa (with some lost to inefficiency), in accordance with conservation laws.
- Having a device drip for a long time is meaningless if you're extracting very little energy from it. See: http://www.emoti.com...ag/00/0417.html Tar can continue dripping extremely slowly for a century but no one would consider it a perpetual motion machine. The longer your machine runs, the easier you're making it to fool yourself (and possibly others).
- You have not shown how or even speculated that your machine circumvents conservation of energy laws. There is no reason to believe it does, other than that it's a puzzle to figure it out.

On the plus side: You may consider this device among the best "perpetual motion machines" ever designed and built, some of which might be considered "famous".
On the bad side: None have ever worked, and there is scientific proof of that, and a lot of time has been and continues to be wasted.

On a related note... I finally discovered the "Stop watching this topic" button which has improved my life! I keep fooling myself into thinking that I can help resolve something with a reply.

Edited by md65536, 16 February 2011 - 11:06 PM.

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### #50 christopherkirkreves

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Posted 20 February 2011 - 09:50 PM

michel123456

Thank you. I read through the wikipedia link. I don't know anything about magnetorheological fluids other than what I just read. And I think I know what you're suggesting, but I'm not exactly sure.

Let me try

1. Magnetorheological fluids and ferrofluids are smart fluids.

2. When smart fluids are exposed to a magnetic field their viscosity increases.

3. When a fluid increases in viscosity it becomes more "quasi-solid".

4. The greater the magnetic field, the greater the increase in viscosity.

5. However, at some point a smart fluid will become "magnetically saturated," meaning additional increases in the magnetic field will not increase the apparent viscosity of the fluid.

6. In the machines I built, the ferrofluids were exposed the a magnetic field for several days, and over those several days they become more and more viscous, eventually reaching a point where they were brittle and solid like.

Therefore

7. Perhaps the reason why my ferrofluids became brittle and stopped moving was not evaporation (or not just evaporation) but also "magnetic saturation."

8. And, while the evaporation problem is something that can be removed in a better built machine, given the nature of the design (specifically that the ferrofluids are exposed to a magnetic field and left there), that "magnetic saturation" will always eventually be reached, and so the ferrofluid will always eventually become brittle, and so these machines will always necessarily come to an eventual stop.

9. And, thus, it is a failed design.

Is this the argument? If so, it's a good one.

But, before I respond to this, and I think there is a response, I'd like to understand if I've gotten the argument right.

michel123456, did it get the argument and analysis right?

Thank you.

md65536

"Evaporation typically involves an energy input."

And it is an essential part in the movement of water though a tree. However, here it does not aid in the movement. (I did get one machine to drip, and then submerged it in water. It continued to drip for 10 days.)

"Conservation of energy is what you need to research."

"I think that it is very unlikely that you can break a law of science without first understanding it enough to know where (and only IF) it can be broken."

Are you suggesting that laws of physics are not "falsifiable"? If so, then you are suggesting, it appears, that they are not scientific theories. They're more like religious principles.

It used to be thought that the hallmark of a scientific theory was whether or not the theory was "testable." So, for example, if my theory is that "god exists" but there's no way to test this, then it's not a scientific theory. But, they later realized that it was not enough to be able to test an idea, but rather to be able to show it is false. So, for example, if my theory is "if I say something mean to you, you will be insulted," and this (you being insulted) is demonstrated by you saying something back to me, or walking away, or saying nothing, or whatever else you might possibly do, then if any reaction on your part is evidence of you being insulted, then while we can test my theory, we cannot falsify it. And, therefore, it is not a scientific theory. Rather, a scientific theory (in my understanding) looks more like "if I let go of this ball it will fall." While it might be true that every time I drop this ball, over the course of my lifetime, it falls, the fact that it is in the realm of possibilities that it could do something else (like float up), means that it is possible to show my theory false, and therefore it is a "scientific theory."

(Again, if I understand the philosophy of science correctly in this regard.)

Are you are saying that I need to go think about the law of "Conservation of Energy" and specifically whether or not "IF" it can be broken? If this law is not falsifiable then, in my current understanding, it is not a "scientific theory."

And, if it is possible (though very unlikely) to show this law to be false, then what would that demonstration look like? It would look like a mechanism, or some other system, where energy is not conserved: a perpetual motion machine, or some other possible (though very unlikely) natural phenomena.

"A pendulum won't swing to the bottom and then immediately stop, "

"Tar can continue dripping extremely slowly for a century "

My claim that there is a conservation of energy issue in this design is not based on these machines dripping for several days. (I'll go you one further. Pick up a rock, put it on top of a rocket, and shoot it off into free space where there aren't many stars. This rock will move for a longer time than a pendulum or tar.) Rather, my claim of a conservation of energy issue is that it appears ( although I'm open to suggestions of possible problems like "magnetic saturation"  ) that there is an issue after each capillary drips once. After each capillary drips once (and friction is encountered) and as the fluid moves back to an equivalent position to where it began, energy (thermal energy) is increased without an offsetting decrease another form of energy. (Just as a hypothetical pendulum, which encounters friction while it swings, but then still manages to reach the same height on the other side.)

" which has improved my life!"

I'm sorry this experience has been an unpleasant one for you.

Thank you  regardless.

And  michel123456, let me know if I got the "magnetic saturation" argument and analysis right? I think I know the right response, but it's challenging.

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### #51 md65536

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Posted 21 February 2011 - 02:33 AM

"Evaporation typically involves an energy input."

And it is an essential part in the movement of water though a tree. However, here it does not aid in the movement. (I did get one machine to drip, and then submerged it in water. It continued to drip for 10 days.)

Evaporation can use energy to extract liquid from a capillary, freeing it to draw up more liquid. If your device is indeed cycling through the same energy states (which I don't think it is but it could be), this could allow it to do so.

"I think that it is very unlikely that you can break a law of science without first understanding it enough to know where (and only IF) it can be broken."

What I meant by this... perhaps I should clarify -- I'll restate it:

It is VERY UNLIKELY that you can
1. set a goal of breaking a well-established law of science,
2. undertake that goal without understanding the law you're trying to break, and
3. succeed.

Or in other words: It's very unlikely that you can design a device specifically to break laws that you don't understand, and have it work.

PERHAPS if you understood the law and came up with a theoretical way to circumvent it (ie. to falsify it),
MOST LIKELY you'd still be wrong, but you might not be. It happens and it will continue to happen... it is part of scientific process.
However, laws that are confirmed by hundreds of years of experimental confirmation tend to very rarely be false.

PERHAPS if you already HAD a device which SEEMS to break the laws of physics (say if you had a device that ran on 2 car batteries and unexpectedly kept running after you removed them),
even then it is very unlikely that it breaks existing laws of science, or requires new ones. It is more likely that there is some explanation that you're missing.

I don't want to discourage you from trying to do the impossible, but there's a fine line between attempting something new without knowing exactly what you're doing, and devoting your time to a hopeless project (like a PMM) while refusing to understand it. (To your credit, you are attempting to understand it, despite refusal to accept the key principles that would let you understand it.)

Yes, conservation laws are falsifiable. Your device could be considered an experiment which supports the well-known conclusion: They are not false.

I'm sorry this experience has been an unpleasant one for you.

No, the apologies are mine. Discussion can always be beneficial though my negativity may not. And you seem to be committed to truly understanding your device; it should not be my concern what route you take to get there.

Edited by md65536, 21 February 2011 - 03:09 AM.

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### #52 michel123456

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Posted 21 February 2011 - 06:54 PM

michel123456, did it get the argument and analysis right?

You can call me Michel.
123456 is a code.

Yes. But I am not so passionate in your device as you are, and I didn't go into a very profound cogitation. Basically, my argumentation ends with the "maybe this" comment.

Edited by michel123456, 21 February 2011 - 07:44 PM.

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Michel what have you done?

### #53 christopherkirkreves

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Posted 24 February 2011 - 12:23 AM

md65536:

"No, the apologies are mine."

When I go to the supermarket and stand in line and see a tabloid magazine with an article like "An Alien Ate My Baby," I think to myself, perhaps rightly or wrongly, "How stupid it is that people actually believe this."

I understand that for me to post a "Perpetual Motion Machine" design in a forum such as Science Forums, where the people here are educated in science, that the reaction I will realistically get will be similar to my feelings in the supermarket.

And I understand that it further might even seem incredulous that the person proposing this design, which if true would disprove a Law of physics that has been understood and verified over hundreds of years, has taught himself (and is teaching himself with whatever help he can get) physics, and makes mistakes ( but then does learn from them and corrects them ) like, one, talking about the empty spaces in these capillaries after they've been saturated, and like, two, talking about potential energy being created and destroyed as a dry capillary is created and destroyed.

Your attitude towards me is very understandable.

"Evaporation can use energy to extract liquid from a capillary, freeing it to draw up more liquid."

After the capillary becomes saturated, the fluid spikes out beyond the end of the capillary. This spike is horizontal to the force of gravity. This spike is inline with the lines of flux between the two magnets. This spike then grows horizontally and inline with the lines of flux. And, when the spike breaks off from the end of the capillary and drips, it drips horizontally, inline with the lines of flux, and onto the face of the second magnet.

I agree that evaporation can be the force to pull a fluid from a capillary. But it does not seem to be the force at work here. The force at work here, pulling the fluid out from the capillary, seems to be magnetic attraction.

"It's very unlikely that you can design a device specifically to break laws that you don't understand, and have it work."

I think two things need to be separated out here:

1. Understanding the First Law of Thermodynamics.

2. Understanding the mechanics of my design.

My understanding of the First Law of Thermodynamics is this: while energy can change from one form to another, there is never an increase nor decrease in the overall quantity of energy (in a closed system). If we make our closed system the entire Universe, then, in theory, we could go around and add up all the energy, in all its various forms, and come up with a number (e.g. 1.25 zillion units of energy). And, if we then wait five minutes, or a thousand years, and add up all the energy in the Universe again, we will probably find different amounts of energy in it's different forms (e.g. thermal, kinetic, potential), but we will get the exact same total number (1.25 zillion units).

To make it even more simplistic, at Time One a ball has just left my hand and is rolling across the floor. This motion is 10 units of kinetic energy. We add this 10 units along with all the other forms of energy in the Universe to get a total number: 1.25 zillion units. This ball will come to a stop. At Time Two when we add up all the energy in the Universe, this 10 units of kinetic energy is gone. However, what brought the ball to a stop is friction. And friction produces heat. And heat is another form of energy. So, at Time Two, while there is a decrease in 10 units of kinetic energy, there has been an increase in 10 units of thermal energy, so the overall amount of energy within the Universe is the same.

This is my understanding of The First Law of Thermodynamics. Did I get it right?

Now, when it comes to the mechanics of my pmm design, I think I understand them. And if I do, then there is a conservation of energy issue. And, if there is not a conservation of energy issue, then I've misunderstood the mechanics. It has been suggested that I'm ignoring the reasons given to me why this pmm design will not work. I don't think I have. And I hope not.

"Your device could be considered an experiment which supports the well-known conclusion "

In the philosophy of science you can never technically prove a theory true (but you can prove a theory false). However, you can show a theory is more and more and more and more likely to be true. This is not my first pmm design. All the proceeding ones I can explain why they fail. I can explain why/how energy is conserved. And I'd like to think I've added (along with everyone else in the last few hundred years) to showing that the First Law of Thermodynamics is more and more and more likely to be true.

However, I have yet to find a way out of the conservation of energy issue in this design. Maybe, as it has been suggested, it has been shown to me in this thread, but I'm ignoring it. I hope not.

Thank you.

michel123456 (aka Michel):

"Yes. But I am not so passionate in your device as you are, and I didn't go into a very profound cogitation. Basically, my argumentation ends with the "maybe this" comment."

Yes. I understand you have more of a "passive interest" in my device, as compared to my "passionate interest," and you were merely suggesting "maybe this" is an area for me to look into. I did. I've studied the wikipedia article, and with the hope that what's posted there is true:

1. "Magnetic saturation" occurs due to the intensity of the magnetic field, and not time spent in the magnetic field. A magnetorheological fluid will become more and more solid like as the intensity of the magnetic field increases. However, it reaches a point where increases in the intensity of the magnetic field will not be followed by increases in the solidity of the magnetorheological fluid. "Magnetic saturation" is dependent on the intensity of the field and not time in the field. (If I understood the article right.) The ferrofluids in my machines work just fine for the first few days. The intensity of the magnetic fields do not make them solid like and motionless. It's only after a few days that this happens (and there is no change in the intensity of the magnetic fields over those days).

2. However, according to wikipedia, there is "Particle Sedimentation" in a magnetorheological fluid: "Ferroparticles settle out of the suspension over time due to the inherent density difference between the particles and their carrier fluid." (Which is probably what I should have talked about in my last post where I tried to detail out your "maybe this" argument.)

3. However, this is where the differences between magnetorheological fluids and ferrofluids becomes important. According the wikipedia they are basically the same thing, but just that magnetorheological fluids have larger suspended particles("micrometre-scale"), while ferrofluids have smaller particles("nanoparticles"). But this difference is size leads to a real difference between them, in that magnetorheological fluids are not subject to Brownian motion, while ferrofluids are. (wikipedia: "Ferrofluid particles are primarily nanoparticles that are suspended by Brownian Motion and generally will not settle under normal conditions.")

4. And, also according to wikipedia, there are two ways to slow down the particle sedimentation of a magnetorheological fluid: one, the addition of a surfactant, and, two, the addition of ferrofluid sized nanoparticles:

"Surfactant-aided prolonged settling is typically achieved in one of two ways: by addition of surfactants, and by addition of spherical ferromagnetic nanoparticles. Addition of the nanoparticles results in the larger particles staying suspended longer since to the non-settling nanoparticles interfere with the settling of the larger micrometre-scale particles due to Brownian motion."

I think the key phrase in the above quote is " the non-settling nanoparticles ".

5. So, I'm new the "smart fluids" and their workings, and there's a lot more I need to learn. But it doesn't look like ferrofluids will corrupt in these machines due to "magnetic saturation" or "particle sedimentation". But I'm open to finding out why I'm wrong.

Thank you.

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### #54 Ethereally Luminous

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Posted 2 March 2011 - 01:44 AM

If all the poles are south why would the magnetic fluid want to travel at all? Do the particles in the fluid exhibit the same form of polarity as a standard magnet?
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### #55 christopherkirkreves

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Posted 3 March 2011 - 10:13 PM

Ethereally Luminous

"If all the poles are south why would the magnetic fluid want to travel at all?"

Thank you. In my first post in this thread I used this drawing but had the poles on the second wedge magnet backwards. The North side of one wedge faces the South side of the other wedge (and I had, as you pointed out, two South sides facing each other). Thank you.

"Do the particles in the fluid exhibit the same form of polarity as a standard magnet?"

My understanding is that the particles in a ferrofluid are like a metal ball bearing. When the ball bearing is not in a magnetic field it has no polarity. However, when it is in a magnetic field, it's polarity is in line with the lines of flux from the magnetic running through it. And when it's removed from the magnetic field it will eventually lose its polarity. It's my understanding that the particles in the ferrofluid work the same way. But you may be asking a more sophisticated question than I have the answer to.

Thank you.
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### #56 christopherkirkreves

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Posted 31 March 2012 - 04:38 PM

after some much needed librarial andkerouacian studies, this question

A Potential Energy and Capillary Action Question

As fluid moves up a capillary, gravitational potentialenergy is increased.

Energy is conserved because this increase is offset by adecrease in the potential energy due to surface tension.

If there is a 10 unit increase in gravitational potentialenergy, then there is a 10 unit decrease in the potential energy due to surfacetension.

If a bent capillary has the same length and the same innerdiameter as the straight capillary, then the same amount of fluid moves intothe bent capillary as moved into the straight capillary.

There is the same amount of contact between the fluid andthe capillary walls. There is the sameamount of adhesion.

There is the same amount of change in surface area. There is the same amount of decrease insurface tension.

If the decrease in potential energy due to surface tension is 10 units with the straight capillary, then there is also a 10 unit decrease inpotential energy due to surface tension with the bent capillary.

The increase in gravitational potential energy is less inthe bent capillary than in the straight capillary.

If there is a 10 unit increase in gravitational potentialenergy with the straight capillary, then there is a < 10 unit increase ingravitational potential energy with the bent capillary.

How then is energy conserved?

---

Is the Question Moot?

1.

>> The question ignores the "point of equilibrium."

<< The top of the straight capillary (and therefore thetop of the bent capillary) is lower than the point of equilibrium. Fluid reaches the top.

2.

>> The question needs to be put in the inverse where withthe bent capillary there is a 10 unit increase in gravitational potentialenergy.

<< This, then, means there is a > 10 unit increase ingravitation potential energy with the straight capillary. So, the question remains.

---

How then is Energy Conserved?

1.

>> Surface Tension Energy.

<< If the increase in gravitational potential energy witha rising column of fluid in a capillary is offset by an equal decrease in potentialenergy due surface tension then there is a logical problem.

<< The decrease in surface tension is the same with boththe bent and straight capillaries. The decrease in potential energy due tosurface tension is the same with both the bent and straight capillaries. The increase in gravitational potentialenergy is different with the bent and straight capillaries. If the increase in gravitational potentialenergy is offset by a decrease in potential energy due to surface tension, thenthere is a conservation of energy issue.

2.

>> Capillary Potential Energy (Extended Adhesive PotentialEnergy).

<< If the increase in gravitational potential energy witha rising column of fluid in a capillary is offset by an equal decrease incapillary potential energy then there is a logical problem.

<< If the potential energy decrease comes from a potentialenergy within a dry capillary, then when a dry capillary is dismantled, adecrease in energy (it takes energy to dismantle a dry capillary) leads to adecrease in another form of energy, capillary potential energy. The potential energy cannot be within the drycapillary itself.

3.

>> Intermolecular Adhesion and Cohesion Potential Energy.

>> It is not only the extended adhesive potential of thedry capillary used in lifting up the column of fluid against the force ofgravity, but also the interaction between the fluid experiencing adhesion withthe walls, and that same fluid cohering with the rest of the cohering column offluid.

<< If the increase in gravitational potential energy witha rising column of fluid in a capillary is offset by an equal decrease inintermolecular adhesion and cohesion potential energy then there is logicalproblem and an oversight.

<< The adhesion part of this intermolecular mixturerequires the existence of the capillary walls. A dry capillary cannot have any of the potential energy withinitself. If so, then destruction of a drycapillary leads to another conservation of energy issue.

<< The amount of adhesion and cohesion is the same withboth the straight and bent capillaries, so this cannot offset the differencesin increases in gravitational potential energy between the bent and thestraight capillaries.

4.

>> There is more pull on the walls in the straightcapillary than there is pull on the walls in the bent capillary.

<< Force is not energy. Force is not used up.

5.

>> Combination of Fluid and Capillary.

<< The potential energy cannot be held, even partially,within the dry capillary, without creating another conservation of energyissue.

6.

>> Fluid Potential Energy and Capillary Opportunity.

>> The potential energy, and the potential energydecrease, is all within the fluid. Thecapillary gives the fluid the opportunity to decrease this potential energy. The straight capillary has more opportunityfor the fluid to use its potential energy due to surface tension and the bentcapillary has less opportunity for the fluid to use its potential energy due tosurface tension. The greater theopportunity, the greater the decrease in potential energy due to surfacetension.

<< There is the same change in surface tension with thebent and straight capillaries. There isthe same decrease in surface tension with the bent and the straightcapillaries. There is the same decreasein potential energy due to surface tension with the bent and the straightcapillaries.

<< If there is a greater opportunity, it does not happen.

7.

>> Temperature.

>> The proposal: As gravitational potential energyincreases, thermal energy decreases. Andso with the fluid moving higher up in the straight capillary than the fluid inthe bent capillary, the temperature within the fluid in the straight capillarywould be lower than the temperature within the fluid in the bent capillary.

>> There are web sites where they claim decreases in gravitationalpotential energy lead to increases in thermal energy. And they claim to have experiments to provethis.

<< There are not web sites (not that I could find) wherethe claim is made that they have experimental evidence for increases ingravitational potential energy leading to decreases in thermal energy.

<< If the increase in gravitational potential energy witha rising column of fluid in a capillary is offset by an equal decrease intemperature then there is a logical problem.

<< This concept can lead to conservation of energy withcapillary action, but if then applied to other conservation of energy analyses,then, there energy is not conserved.

<< As a pendulum swings up, its kinetic energy is changedinto an equal amount of gravitational potential energy. And as a pendulum swings down, itsgravitational potential energy is changed into an equal amount of kineticenergy. If there is friction, then thisback and forth pool of energy, slowing becomes an equal amount of thermalenergy, as friction creates heat and bring the motion to a stop.

<< If we add the concept "gravitational potential energyincreases correspond to thermal energy decrease /and/ gravitational potentialenergy decreases correspond to thermal energy increases" to the analysis of aswinging pendulum, then energy is not conserved in that there is an decrease inthermal energy as the pendulum swings up and an increase in thermal energy asthe pendulum swings down. Conservationof energy is not variable nor cyclical.

<< And, why would a substance moving up against the forceof gravity due to capillary action have a thermal energy decrease while asubstance moving up against the force of gravity due to something else, such asbeing in motion or being lifted by hand, does not?

8.

>> Depth Pressure.

>> Pressure increases with depth, and the column of fluidis taller in the straight capillary than in the bent capillary.

<< Force is not energy. Force is not used up.

9.

>> Relativity.

<< If Relativity is used to find differences in anotherform of energy to compensate for the differences in increases in gravitationalpotential energy between the bent and straight capillaries, then that logicmost likely will lead to other conservation of energy issues when appliedelsewhere without capillary action, such as with a swinging pendulum.

10.

>> Time Spent Moving In the Gravitational Field.

>> It takes longer for the fluid to move against the forceof gravity to the top of the straight capillary than to the top of the bentcapillary.

<< If the time spent moving against the gravitationalfield leads to differences in some form of energy between the straight and bentcapillaries, then this will lead to conservation of energy issues when this isapplied to other conservation of energy analyses, such as a swinging pendulum.

11.

>> Distance Moved In the Gravitational Field (On A MicroLevel).

>> The fluid also moves farther against the force ofgravity in the straight capillary than the fluid moves against the force ofgravity in the bent capillary.

<< The further an object moves against the force ofgravity, the greater the potential energy. This is true with capillary action, swinging pendulums, and anythingelse. And there is an equal decrease inanother form of energy, like kinetic energy, to offset this increase.

<< However, there is not a decrease in something likethermal energy with an increase gravitational potential energy. And there is not something like a decrease inthermal energy that is greater or lesser depending on how far an object movesagainst the force of gravity.

<< A micro level look at "distance moved in thegravitational field" does not lead to an offsetting balance to the differencesin increases in gravitational potential energy between the bent and thestraight capillaries.

12.

>> Velocity and Collision.

>> The fluid in the bent capillary, presumably, reaches agreater velocity than the fluid in the straight capillary because the fluid inthe straight capillary has to move more against the force of gravity.

>> When the two columns of moving fluid come to a stop,there is a greater impact with the bent capillary than there is with thestraight capillary, because this fluid is moving faster.

>> The greater the impact the greater the thermal energyincrease. More thermal energy is createdwith the fluid in the bent capillary than is created with the fluid in thestraight capillary. This differenceoffsets the difference in increases in gravitational potential energy.

<< Where does the kinetic energy come from?

<< If it comes from the potential energy due to surfacetension, then two different amounts of kinetic energy come from equal amountsof potential energy due to surface tension, and this is a conservation ofenergy issue. If it comes from thecapillary in its shape size or position, fully or in part, then there is aconservation of energy issue when a dry capillary is dismantled.

<< Noting that the difference in gravitational potentialenergy is preceded by a difference in kinetic energy gets us out of one conservationof energy issue, potentially, but then, definitely, into another one.

---

Other Considerations?

1.

>> The concept of "work" is not addressed.

<< The intermolecular forces of adhesion and cohesion actover the same distance in both the bent and straight capillaries.

2.

>> The concept of "conservation of momentum" is notaddress.

<< There does not seems to be a link between "conservationof momentum" and a difference in a form of energy to offset the differences ingravitational potential energy between the straight and the bent capillaries.

3.

>> Einstein largely distanced himself from his firstpublished paper where it follows from a linear relationship between temperatureand surface tension that, inter alia, a surface can be considered a potentialenergy.

<< please see"The Centenary of Einstein's First Scientific Paper", J.N. Murrell and N.Grobert, Notes Rec. R. Soc. Lond. 56 (1), 89-94 (2002).

---

How then is energy conserved?

Thank you.

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### #57 christopherkirkreves

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Posted 13 April 2012 - 04:26 PM

"A Potential Energy and Capillary Action Question" Addendum

---

13.

>> The "meniscus" is not addressed.

<< (Again, the tops of the capillaries are below the point of equilibrium. The liquid reaches the tops.)

<< (The attraction between the liquid and the capillary walls is greater than cohesion, while the attraction between the liquid and the reservoir walls is equal to or less than cohesion.)

>> A larger meniscus means less of an increase in gravitational potential energy.

>> If the meniscus with the straight capillary is larger than the meniscus with the bent capillary then the increases in gravitational potential energy are closer to being equal.

<< However, the meniscus is part of the surface area.

<< A larger meniscus means less of a decrease in surface area.

<< Less of a decrease in surface area means less of a decrease in the potential energy due to surface tension.

<< If the meniscus is larger with the straight capillary than with the bent capillary, then the decreases in the potential energy due to surface tension are no longer equal.

<< The closer the two are in gravitational potential energy, the greater the difference is between the two in decreases in potential energy due to surface tension.

<< A larger meniscus with the straight capillary does not resolve the conservation of energy formula.

<< (This seems like a possible solution to the conservation of energy formula if we think of wide capillaries without much difference in height. But if the width of the capillaries is space between two glass plates pressed together, and if the height of the bent capillary is fractions of an inch, whereas the height of the straight capillary is several feet, then it's hard to imagine slight differences in the meniscuses equaling out the difference in increases in gravitational potential energy between the bent and the straight capillaries (even if the then created difference in the decreases in potential energies due to surface tension is ignored).)

How then is energy conserved?

Thank you.

Edited by christopherkirkreves, 13 April 2012 - 05:56 PM.

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### #58 christopherkirkreves

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Posted 20 April 2012 - 07:35 PM

I've been forwarded several answers to this question. They've all varied wildly, but none of them have seemed to resolve the conservation of energy formula. That is, until yesterday. Yesterday, I was forwarded one that works.

The 10 unit decrease in potential energy due to surface tension does not become a 10 unit increase in gravitational potential energy with the straight capillary.

Rather, with both the straight and the bent capillaries, the 10 unit decrease in potential energy due to surface tension becomes a 10 unit increase in the sum of gravitational potential energy plus kinetic energy.

The more the fluid moves against the force of gravity, on its way to the ends of the capillaries, the slower it will move.

In the straight capillary the fluid moves further against the force of gravity. The fluid moves higher, so a greater increase in gravitational potential energy. The fluid moves slower, so a lesser increase in kinetic energy.

In the bent capillary the fluid moves less against the force of gravity. The fluid moves up less, so a lesser increase in gravitational potential energy. But, the fluid moves faster, so a greater increase in kinetic energy.

(And when the motions come to a stop, the differences in kinetic energy become differences in thermal energy.)

Thus, energy is conserved (and the flaw in my logic is exposed).

Another Question

Two identical capillaries are in two cups of water.
One cup of water is narrow, and the other cup of water is wide.

Water moves up each capillary. The water moves up the same height. However, the water level in the narrow cup drops more than the water level in the wide cup. There is a greater increase in gravitational potential energy with the narrow cup than with the wide cup.

Surface Changes:

1. The liquid-air contact remains unchanged with both the narrow and wide cups.

2. The liquid-inner_capillary_walls contact increase equally with both the narrow and wide cups.

3. The liquid-cup&outer_capillary_walls contact decreases more with the narrow cup than with the wide cup.

The surface tension between the liquid-cup&outer_capillary_walls is more favorable than the surface tension between the air-cup&outer_capillary_walls. There is a change to this less favorable state with both the narrow and wide cups, but more so with the narrow cup. So, there is less of a decrease in the potential energy due to surface tension with the narrow cup than with the wide cup.

Thus, not only is there a greater increase in gravitational potential energy with the narrow cup, but this comes from less of a decrease in the potential energy due to surface tension.

As the liquid is pulled up into the two capillaries, the fact that the water level drops further in the narrow cup than in the wide cup, should not affect the velocity of the lifting liquids. Differences in kinetic energy conceptually balance out the different increases in gravitational potential energy between the bent and straight capillaries, but there doesn't seem to be the same differences in kinetic energy here to balance out the different increases in gravitational potential energy (and the different decreases in the potential energy due to surface tension).

How then is energy conserved?

(1. The capillaries are below the point of equilibrium. Fluid reaches the tops. The meniscus will likely be larger in the capillary with the narrow cup than in the capillary with the wide cup. This lessens the increase in gravitational potential energy (and increases the liquid-air contact). But, if the width of the capillaries is like that when two glass plates are pressed together, and if the height of the capillaries is several feet, then it hard to imagine small differences in the meniscuses balancing out the difference in drops in water levels between a very narrow and a very wide cup.)

(2. There is perhaps an intuitively misleading aspect to the question. The greater drop in water level might seem to suggest that the liquid in the narrow cup puts more upward pressure on the liquid in the capillary and this will somehow lead to greater kinetic energy in the capillary with the narrow cup. But it's the exact opposite. The liquid levels start out at the same height, but the liquid in the narrow cup quickly becomes lower than the liquid in the wide cup. So there is an equal amount of pressure and then less pressure from the liquid in the narrow cup, than from the liquid in the wide cup, on the liquid in the capillaries.)

(3. The liquid that remains in the cups moves as liquid is lifted up and into the capillaries. If the velocity of the rising liquid in the capillaries is the same, then the liquid that remains in the narrow cup moves faster than the liquid that remains in the wide cup. But, more fluid is set in motion in the wide cup than in the narrow cup. So, this movement, it seems, should balance out.)

(4. The less favorable surface tension change in the narrow cup than in the wide cup, if anything, would seem to suggest a lessened velocity increase in the liquid in the narrow cup and thus, if anything, would lead to lessen the velocity of the liquid moving up the capillary from the narrow cup than from the wide cup.)

Thank you.

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### #59 christopherkirkreves

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Posted 12 July 2012 - 08:32 PM

A "Wedge Circle" and Brownian Motion Question

side view

A small solid is floating on the top of a liquid. It is contained within barrier walls. The barrier walls reside above the liquid and do no directly interact with the fluid.

The small floating solid is subject to Brownian Motion. It moves randomly. And, from time to time, it randomly collides with the barrier walls.

top down view

The shape of the barrier is a "wedge circle." The widest part of the wedge is closed off with a straight wall. The smallest part of the wedge is open to the circle.

When the small solid collides with the wedged wall, randomly over and over again, the overall angle of reflection is counter clockwise. When the small solid collides with the straight wall, randomly over and over again, the overall angle of reflection is clockwise.

top down view

The small solid and wedged wall collisions are elastic. The small solid and straight wall collisions are inelastic. After colliding with the wedged wall the kinetic energy of the small solid is preserved in its overall counter clockwise reflection. While, after colliding with the straight wall the kinetic energy of the small solid is not preserved in an overall clockwise reflection.

top down view

So, over an extended period of time, the small solid will move in an overall counter clockwise circular path.

When the small solid moves in an overall counter clockwise circular path, and does so over and over again, is "work" being done?

---

1. A Feynman Brownian Ratchet fails to produce rotation in one overall direction because larger and smaller fluctuations in Brownian Motion in the fluid and solid elements of the system push and retard the ratchet "forwards" and "backwards" equally. While in the question presented here, the Brownian Motion in the small floating solid and in the solid barrier walls will randomly make the reflections more or less clockwise and more or less counter clockwise. With these variations in the angles of reflection being random, however, the probability is that overall they will have no net effect. This question is not just another version of the Feynman Brownian Ratchet. Here, there is a small solid having elastic and inelastic collisions with angled solid barrier walls.

2. The collisions do not have to be perfectly elastic or inelastic. If the collisions with wedged wall are more elastic and if the collisions with straight wall are more inelastic, an overall counter clockwise circular path is still created, and the question remains.

3. If the space above the liquid is filled with air, then the Brownian Motion interaction between the gas and the barrier walls can complicate the analysis. This is eliminated if we assume a vacuum and not air in the space above the liquid.

4. When the small solid collides with the inner circular wall, randomly over and over again, the overall angle of reflection from this is neither more clockwise nor more counter clockwise.

5. It would be more accurate to say "When the small solid collides with the wedged wall, randomly over and over again, the probability is for the overall angle of reflection to be counter clockwise." And, it would be more accurate to say "When the small solid collides with the straight wall, randomly over and over again, the probability is for the overall angle of reflection to be clockwise." So, if the collisions with the wedged wall are elastic and the collisions with the straight wall are inelastic, the question should then be "When, as is probable, the small solid moves in an overall counter clockwise circular path, and does so over and over again, is "work" being done?"

6. If the collisions with the straight wall are perfectly inelastic, from time to time, the small solid will come to a stop and its kinetic energy will be returned to thermal energy. If periodically stopping seems problematic, then the straight wall can be removed and replaced by an outer circular barrier wall where the collisions will be elastic. Now, the small solid will move in an overall counter clockwise path while within the inner "wedge circle" and will move in an overall clockwise path while within the outer "wedge circle."

7. If "work" is being done, what are the second law implications?

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### #60 christopherkirkreves

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Posted 5 September 2012 - 11:25 PM

Another "Wedge Circle" and Brownian Motion Question

---

side view / top down view

A small solid is floating on the top of a liquid.
It is contained within barrier walls. The barrier walls reside above the liquid and do not directly interact with the fluid.

The shape of the barrier is a "wedge." There are three walls: a long wedged barrier wall, a short straight barrier wall, and a long straight barrier wall. There is a small opening at the narrow end of the wedge, and a small opening at the wide end of the wedge. The barrier walls are fixed in place and do not move relative to the container of fluid.

This is all within a closed system.

top down view

The small floating solid is subject to Brownian Motion. It moves randomly. And, from time to time, it randomly collides with the barrier walls.

When the small floating solid collides with the barrier walls, randomly over and over again, the overall angle of reflection is perpendicular to the angle of each wall.

The overall angle of reflection from colliding with the long wedged barrier wall is slightly to the left. The overall angle of reflection from colliding with the short straight barrier wall is directly to the right. And the overall angle of reflection from colliding with the long straight barrier wall is neither more left nor more right.

(There is also Brownian Motion within the small floating solid and within the solid barrier walls. These fluctuations will randomly make the angles of reflections greater and smaller than the angles of incidences by varying degrees. However, since these fluctuations are random, the probability is that there will be no overall net effect. The overall angle of reflection will be perpendicular to the angle of each barrier wall.)

top down view

There is an area around the small floating solid, within which, it will likely move to next. The specific place it will move to next within this area is random (with the exception that the small floating solid will more likely move a shorter rather than a longer distance). If the distance between the long wedged barrier wall and long straight barrier wall is within this area, then the reflections off of these walls will significantly interact with each other, and after reflecting off of one of these walls the small floating solid will either move out into the open space or it will cross the open space and collide with the other wall.

top down view

If the small floating solid moves out into the open space, and then comes to a stop, its next move will be random. However, with the long wedged barrier wall and the long straight barrier wall being within this area, the movement of the small floating solid will not be totally random. The trajectories of the overall reflections off of these walls crisscross with each other, and nudge the small floating solid more and more to the left in an overall zigzag pattern.

(The actual path of the small floating solid will not strictly be a zigzag. The actual path of the small floating solid will be filled with random movements in all directions. However, overtime, the small floating solid will be nudged more and more to the left in an overall zigzag pattern.)

If the small floating solid crosses the open space and collides with other wall, it will then be reflected off of the other wall. The overall angle of reflection off of the long wedged barrier wall is slightly to the left. So when the small floating solid reflects off of the long wedged barrier wall and then crosses the open space and collides with the long straight barrier wall, overall, the small floating solid will be moving slightly more to the left. And thus the overall angle of reflection off of the long straight barrier wall will no longer be neither more left nor more right. The overall angle of reflection off of the long straight barrier wall will be slightly left. And this will make the overall zigzag pattern slightly more pronounced to the left.

(This more pronounced zigzag pattern also comes from the overall neither more left nor more right longer reflections off of the long straight barrier wall that cross the open space and collide and then reflect off of the angled long wedge barrier wall.)

top down view

When the small floating solid collides with the short straight barrier wall, randomly over and over again, the overall angle of reflection is directly to the right.

At times, the small floating solid will be reflected directly out into the open area of the wedge just to the right of the short straight barrier wall. At other times, after being reflected off ofthe short straight barrier wall, it will collide first with another barrier wall before then being reflected out into the open area just to the right of the short straight barrier wall.

Once in the open area of the wedge, the next movement of the small floating solid will be random. It can move left and collide again with the short straight barrier wall which will reflect it back to the right. It can move to another place within the open area. Or, it can collide with the long wedged barrier wall or long straight barrier wall, where the interaction between these two will start nudging it, overall, back to the left.

top down view

Due to the differences in the reflections from the collisions with the barrier walls and their interactions, the small floating solid will move overall to the left. Then, at some point, the probability is that it will exit the wedge barrier through the small opening on the left.

top down view

If there is an infinitely long line of connected wedge barriers fixed in place above an endless container of fluid, then the small floating solid will move overall to the left forever.

One form of the Second Law of Thermodynamics is "No cycle is possible whose sole result is the abstraction of heat from a single reservoir and the performance of an equivalent amount of work."

"Work" is forcetimes distance. When an object is set in motion "work" is done on that object.

To move the small floating solid continuously overall to the left "work" must be done on the small floating solid. And the "work" done here comes from a single heat bath.

---

top down view

The wedge barrier can be curved around into a "wedge circle." The widest part of the wedge is closed off by a straight barrier wall. There is a small opening between the widest part of the wedge and the narrowest part of the wedge, making it an open circle.

When the small floating solid collides with the outer wedged barrier wall, randomly over and over again, the overall angle of reflection is slightly counterclockwise. When the small floating solid collides withthe straight barrier wall, randomly over and over again, the overall angle of reflection is directly clockwise. And, when the small floating solid collides with the inner circular barrier wall, randomly over and over again, the overall angle of reflection is neither more counterclockwise nor more clockwise.

top down view

The trajectories of the overall reflections off of the outer wedged barrier wall and inner circular barrier wall crisscross with each other and an overall zigzag pattern will occur. This will nudge the small floating solid more and more counterclockwise. The longer overall counterclockwise reflections off of the outer wedged barrier wall that cross the open area and collide with the inner circularbarrier wall will make the overall reflections off of the inner circular barrier wall slightly more counterclockwise, and a more pronounced overall zigzag pattern counterclockwise will emerge.

(This more pronounced zigzag pattern will also come from the overall neither more counterclockwise nor more clockwise longer reflections off of the inner circular barrier wall that cross the open space and collide and then reflect off of the angled outer wedged barrier wall.)

After colliding with the straight barrier wall, at times, the small floating solid will be reflected directly clockwise out into the open space of the "wedge circle" just clockwise of the straight barrier wall. At other times, after colliding with the straight barrier wall, the small floating solid will collide first with another barrier wall before then being reflected out into the open space just clockwise of the straight barrier wall.

Once in the open area just clockwise of the straight barrier wall the next movement of the small floating solid will be random. It can move counterclockwise and collide with the straight barrier wall again which will reflect it back clockwise. It can move to another place within the open area. Or, it can collide with the outer wedged barrier wall or inner circular barrier wall, where the interaction between these two will start nudging it, overall, back counterclockwise.

top down view

The small floating solid will move in an overall counterclockwise circular path. And then, at some point, the probability is that it will move counterclockwise through the small opening from the widest to the narrowest part of the wedge. The motion in an overall counterclockwise circular path will continue. And this process will continue forever.

Just as "work" is done in moving the small floating solid leftward in the infinite line of connected straight wedge barriers, so to "work" is done in the "wedge circle" in moving the small floating solid counterclockwise over and over again. And, here too, this "work" comes from a single heat bath.

If the small floating solid in the "wedge circle" is connected to a round, and very very light weight, magnet, and if this magnet is capable of rotating around a central axis, then this magnet will be rotated overall counterclockwise. And if a wire is placed perpendicular to the lines of magnetic flux, then a current will be created in the wire in one overall direction. And, in accordance with Lenz Law, when the magnet is creating a current in the wire the motion of the rotating magnet (and the connected small floating solid) will be slowed to equal the current created in the wire. The thermal energy of the fluid is turned into the kinetic energy of the small floating solid (and rotating magnet) and this is then turned an electrical current.

Another form of the Second Law of Thermodynamics is "The entropy of a closed system cannot decrease with time."

As long as there is thermal energy the small floating solid (and the magnet) will continue move overall counterclockwise, and more and more thermal energy will be turned into an electrical current.

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What are the Second Law implications of the "wedge circle?"

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1. The Feynman Brownian Ratchet fails to produce rotation in one overall direction because larger and smaller fluctuations in Brownian Motion in the solid and fluid elements of the system push and retard the ratchet "forwards" and "backwards" equally. The question presented here involves a small floating solid having collisions with differently angled barrier walls. The Brownian Motion fluctuations in the solid elements here will make the angles of reflections greater and smaller than the angles of incidences to varying degrees, however, since these fluctuations are random, overall they will have no net effect. The "wedge circle" is not just another version of the Feynman Brownian Ratchet.

2. In the straight wedge barrier, the overall reflections near the corner between the short straight barrier wall and the long wedged barrier wall are largely perpendicular to each wall, even though each wall blocks the open area to one side of the other wall, since that also near the corner between the two walls there will be more reflections off of one wall that then reflect off the other. (And the same is true in the "wedge circle" with the reflections near the corner between the straight barrier wall and outer wedged barrier wall.)

3. There is an overall zigzag pattern that emerges between the short straight barrier wall and long wedged barrier wall in the straight wedge barrier, and between the straight barrier wall and outer wedged barrier wall in the "wedge circle." While moving the small floating solid away from the corner, the small floating solid is moved a little bit more rightward and a little bit more clockwise. While this slows the overall leftward motion and overall counterclockwise motion some, this does not change the overall effects.

4. The analysis is easier if the collisions are considered perfectly elastic. If less than perfectly elastic collisions occur, heat is generated (and the motion of the reflected small floating solid is slowed down). This heat is concentrated, at first, at the point of contact. It will then disperse out from there until it is evenly spread throughout the system. Considering less than perfectly elastic collisions complicates the analysis, but does not change, overall, the angle of reflection. Before the heat is evenly dispersed throughout the system, there will be a greater concentration of heat in the barrier wall, in barrier wall side of the small floating solid, and in the liquid around and beneath the barrier wall. This greater concentration of heat will mean more Brownian Motion activity on the barrier wall side of the small floating solid than on the other. This will push the small floating solid further away from the barrier wall but, overall, it will not change the angle of reflection.

5. The small floating solid does not need to move infinitely leftward or counterclockwise forever for "work" to be done. When Brownian Motion sets a small floating solid in motion, without the presence of barrier walls, in one random direction and then another, "work" is being done. Another form of the Second Law of Thermodynamics is "If a system undergoes spontaneous change, it will change in such a way that its Entropy will increase or, at best, remain constant." It is well known that when Brownian Motion sets a small floating solid in motion (or even when the random fluctuations of Brownian Motion lead to momentary decreases in Entropy within the fluid itself) that this formulation of the Second Law of Thermodynamics is temporarily violated. This is also true when the small floating solid is within the wedge barriers. However, when the small floating solid is within the wedge barriers and the motion of the small floating solid starts and stops and starts again, the overall path of that motion is predictable. There is a greater decrease in disorder when the small floating solid moves around within the wedge barriers than when it moves around without them. And there is a greater decrease in Entropy.

6. There is probably no need for the small floating solid, and the same effects could come from just a fluid in a "wedge circle" shaped container, but the overall movement of a mass of fluid, as opposed to the movement of a single floating solid, is much harder to conceptualize.

Edited by christopherkirkreves, 5 September 2012 - 11:32 PM.

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