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LZJ
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Re: Ellipsoid Power Generation
« Reply #25 on: Apr 12th, 2003, 5:18am » |
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Err...I've already mentioned that in my previous post...
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James Fingas
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Re: Ellipsoid Power Generation
« Reply #26 on: Apr 15th, 2003, 5:46am » |
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Sorry I've been gone from this thread for so long. I have three rebuttals for you, Icarus.
1) You say that your assumption of zero-volume light-emitting objects is more physically valid than my assumption of zero-temperature heat sinks. How is this so? Besides the fact that mine specifically has a law against it, and yours doesn't, of course...
2) Consider the following apparatus: construct two spheres, mirrored inside. One has internal radius A, and the other has internal radius B (A < B). Join the two spheres so that the distance from center to center is just slightly smaller than A+B. Then there will be a pinhole joining the interiors of the two spheres. Put a small ball in the center of each sphere, so that all light going through the pinhole is parallel. Now the amount of light passing through the pinhole from the ball in A will be larger than the amount of light passing through the pinhole from the ball in B (because ball A is closer to the pinhole). This will make the ball in sphere B become hotter than the ball in sphere A.
In case you don't see what I'm getting at, this is exactly analogous to the ellipsoidal reflector (for a very small pinhole).
3) Working from the other direction, if we assume that the second law of thermodynamics holds, then to avoid any power generators like the ellipsoid or these joined spheres, certain conditions must always hold. Here is my statement (I hope that this is clear enough for you to get the gist):
"Choose any two light-emitting surfaces A and B in space. Then, regardless of any distribution of mirrors, lenses, gravitational fields, etc. the 'optical coupling' from A to B must be the same as the optical coupling from B to A. That is to say, the probability that a light ray emitted from A will hit B is the inverse of the probability that a light ray emitted from B will hit A, and is equal to the ratio of the emitting surface areas of A and B. Because of this property, these two surfaces will approach thermal equilibrium with each other over time."
There is no problem if a single point in space does not satisfy this. It is sufficient that it be satisfied for any surface.
Now here are some more problems I don't have the answer to yet:
4) Consider my apparatus from point 2, and put a lens in the pinhole, constructed so that all light from ball A is focussed onto ball B, and all light from ball B is focussed onto ball A (i.e. ball B is the image of ball A). Now ball B will become hotter than ball A, right?
5) Or how about this one: Consider the ellipsoid power generator, but make the ball in the larger ellipse part a lot heavier, so that gravity will bend the light towards the ball. Now it will capture a greater portion of the other ball's light, and will become hotter than the ball in the smaller ellipse part, won't it?
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BNC
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Re: Ellipsoid Power Generation
« Reply #27 on: Apr 15th, 2003, 10:49am » |
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Point 4: No. Since A<B, to be an image of eachother, ball B must be larger than ball A, thus it emitts more energy => more energy reaches A.
Point 5: Without calculating, I assume that once an object heavy enough to bend light is placed in the generator, the path of many more light rays will be changed.. it will probably cancel out in the math.
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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James Fingas
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Re: Ellipsoid Power Generation
« Reply #28 on: Apr 15th, 2003, 2:07pm » |
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BNC,
Ah, now I understand why I don't work with lenses ...
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mike1102
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Re: Ellipsoid Power Generation
« Reply #29 on: Jun 2nd, 2003, 5:52am » |
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wow..... lots of confusion here about heat, temperature and heat transfer.
This "riddle" is really a heat transfer question. Heat transfer doesn't care about photons or black-body radiation/absorption or optics or most of the subjects brought up in the posts. Heat transfer cares about mass, temperature gradients, specific heat and thermal conductivity (see the heat equation). For heat to flow from one point to another, there must first be a temperature difference between those two points - no delta Temp - no heat flow!
Since we're starting with two identical objects at the same temperature and if the walls of the reflector can't absorb heat, there is no delta T, so there is no heat flow - end of story.
There is also an implication in the statement that heat must radiate uniformly into space - this is not the case in reality. Heat will flow in whatever direction there is a thermal gradient, according to a material's thermal conductivity.
My conclusion: This concept for power generation needs a major re-work. But keep trying...... ya just never know what the future holds.
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towr
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Re: Ellipsoid Power Generation
« Reply #30 on: Jun 2nd, 2003, 7:42am » |
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I think the point of this exercise was not to use the laws of thermodynaics to prove they apply to this special case..
Your story doesn't give any explanation of why there isn't any heat exchance (in this case) when there isn't a temperature difference.
Also when you have to objects of equal temperature that touch than their molecules do exchange kinetic energy = heat, but the exchange is equal in both direction (pretty much) giving a net result of no temprature change.
Here the construction of the problem suggest the exchange cannot be equal in both direction, and we have to figure out why it is nevertheless the case.
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Rodrick Crider
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Re: Ellipsoid Power Generation
« Reply #31 on: Jun 3rd, 2003, 11:15am » |
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I don't know anything about thermodynamics, so forgive me if this is completely wrong, but maybe the photons that are eminating from B and bouncing off of the spherical portion have a high probability of colliding with successive photons coming from B, thus causing those photons to scatter instead of returning to B?
If photons can pass through eachother, then this wouldn't make sense. I assume I am probably wrong since I have never seen two laser beams bounce off eachother. Oh well, just a thought...
Rod
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Icarus
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Re: Ellipsoid Power Generation
« Reply #32 on: Jun 3rd, 2003, 6:07pm » |
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Photons do not collide - they pass through each other. Otherwise sight would be useless because the thick sea of photons we live in would be scattering off each other, destroying the images they carry. Photons interact only with electromagnetically charged matter. Since photons themselves are neutral, they do not interact with each other.
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kalie
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Re: Ellipsoid Power Generation
« Reply #33 on: Jun 3rd, 2003, 9:05pm » |
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mike, i guess your the one whos confused. heat transfer does care about photons and radiation, your claim that these concepts are not involved is only valid for a high school student.
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Rodrick Crider
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Re: Ellipsoid Power Generation
« Reply #34 on: Jun 4th, 2003, 9:57am » |
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Thanks Icarus, the more I thought about I realized that that didn't make much sense...I better stick to the easies...
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Icarus
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Re: Ellipsoid Power Generation
« Reply #35 on: Jun 4th, 2003, 3:28pm » |
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on Jun 4th, 2003, 9:57am, Rodrick Crider wrote:| Thanks Icarus, the more I thought about I realized that that didn't make much sense...I better stick to the easies... |
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Nobody always makes sense. About everyone else who has posted to this thread seems sure that I'm not making sense in it! 
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mike1102
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Re: Ellipsoid Power Generation
« Reply #36 on: Jun 5th, 2003, 5:22am » |
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I thought about it some more...... and I was wrong about heat transfer not caring about black body radiation. It certainly does. Black body radiation is the means by which the sun heats the earth and the earth re-radiates energy to keep from burning up.
But I know of no means to accomplish heat flow without a temperature difference. Just like in an electrical circuit, current will not flow without a potential difference; heat will not flow without a temperature difference.
Now.... back to the ellipsoid. If heat flows from A to B, then heat must also be able to flow from B to A. I don't think there is such a thing as a "heat diode" that would allow heat to flow in one direction and not in the opposite direction.
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Icarus
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Re: Ellipsoid Power Generation
« Reply #37 on: Jun 5th, 2003, 5:07pm » |
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on Jun 5th, 2003, 5:22am, mike1102 wrote:| But I know of no means to accomplish heat flow without a temperature difference. |
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No one else seems to be able to figure it out either (unless this trick really does work  ), that's why the 2nd law of thermodynamics is considered a "law"!
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Rujith de Silva
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Finite size of A and B
« Reply #38 on: Oct 14th, 2003, 9:51am » |
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I agree with Icarus that this puzzle should fail for a "robust"
reason. Here's my idea of the "robust" reason:
Assume that A and B are spheres of equal finite size, but small
relative to the rest of the apparatus. Would EVERY ray from A hit B?
Rays that emerge perpendicular to A's surface (i.e., that are
collinear with A's center) would always strike B, this being a basic
property of an ellipse (except for the arbitrarily small amount that
hit A again).
Now consider a nearly horizontal ray that emerges to the right from
the top of A, i.e., with its path as far away from A's center as
possible. This goes to the right, strikes the far right wall, and
would always hit B. Why? Because the "inaccuracy" caused by its NOT
being from the center of A is reduced by the fact that the far right
wall is closer to B than to A. Think about the angles of reflection.
However, consider a nearly horizontal ray that emerges to the LEFT
from A's top. It strikes the far left wall, and then would MISS B,
because the "inaccuracy" is magnified by the far left wall being
closer to A than to B.
So all rays from A will NOT strike B, because of the effect of the
finite sizes of A and B. Similarly, all rays from B will not hit A,
even ignoring those that hit the central "distortion." This holds
true even if A and B are made arbitrarily small, because what matters
is the RELATIVE sizes of A and B.
(Note that all rays from B that hit the central "distortion" will hit
B again.)
What happens to all those rays from A that miss B, and B's rays that
miss A? They will bounce around within the distorted ellipsoid and
hit A or B again after some number of trips. Eventually, everything
will reach equilibrium when the rays permeating any unit area within
the ellipsoid's space are at the same temperature as A and B (sorry, I
don't know the exact thermodynamic terms). At that point, A and B
will also be in equilibrium at the same temperature. Proving this
requires more mathematics/thermodynamics, but I think the basic
argument is correct, and intuitively persuasive.
- Rujith.
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Icarus
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Re: Ellipsoid Power Generation
« Reply #39 on: Oct 15th, 2003, 7:59pm » |
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That is good solution, Rujith. It is I believe the one that James offered to which I objected as the sole cause, though.
However, recalling some of what I have read about other anti-entropic schemes makes me more doubtful that any other reason may exist. Scientific American once published an article about the 2nd Law that had several versions of Maxwell's demon in it. In particular, it examined some mechanical devices intended to move the hot particles only from one side to the other. They examined each and showed the flaw which would cause them to fail. But rather than deep physical principles that applied widely, what they found were simple mechanical problems that were different for each device, but ALWAYS present, no matter how hard you tried to remove them.
So, perhaps I am wrong about there being more.
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| « Last Edit: Nov 12th, 2003, 6:56pm by Icarus » |
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I wonder: Which is larger
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Rujith de Silva
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Re: Ellipsoid Power Generation
« Reply #40 on: Jan 13th, 2004, 12:10pm » |
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Icarus: sorry, I had seen the discussion on point-sized objects, but
had glazed over the stuff on finite-sized objects. I agree that we're
looking for some simple, universal principle that cracks the problem,
but it's eluded us so far. - Rujith.
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