Cont: Brilliant Light Power Going To Market - Free Energy Generator Part 3

[markie]

What gave it that "orbital velocity"? Certainly not simply its "altitude". Otherwise just throwing something in the air would cause it to orbit. Most here already understand that even if you don't.

By orbital velocity I'm referring to tangental orbital velocity. You may be confusing it with escape velocity of something thrown straight up.

 

[JeanTate]

How odd. Why would you 'urge' me to do something that does not pertain to the orbitsphere?

Call me stupid, but I was actually trying to help you with your communication efforts.

First, your posts have given the impression that you do not understand the key parts of classical physics (mechanics) which Mills insists is all you need, and which underlie the "orbitsphere" concept. My suggestion was for a concrete way to show that this is not so.

Second, if you think what I am suggesting something "that does not pertain to the orbitsphere", then I think this illustrates yet another of your failures of communication.

You see, I thought that an orbitsphere includes a ring (hoop, ribbon), an infinitely thin, charged, perfect circle with the charge distributed uniformly (infinitely many such rings in fact).

But OK; you know the saying about horses, water, and drinking?

 

[JeanTate]

<snip>

But optiongeek is impressed by the "simplicity" of equations (10.42), (10.43), (10.46), (10.47), and (10.48), and is willing to overlook the obvious errors in those equations because they produce results that are only slightly less accurate than a simple cubic polynomial approximation. optiongeek dismisses the fact that those equations are considerably less accurate than quantum mechanics, because (1) optiongeek has been taken in by Mills's rhetorical attack on numerical methods and (2) optiongeek is more than willing to imagine conspiracies that, in optiongeek's imaginary world, would account for the accuracy of calculations based on quantum mechanics.

No, I am not going to disabuse optiongeek of his notions.

my hilite.

<snip>

As for people not understanding that solutions need not be closed form, I remember a frustrating discussion with someone who insisted that the fact that there is no analytical solution to the general three body problem means that Newtonian mechanics is flawed (“wrong” was the word I think he used). Of course there are limits to the applicability of Newtonian mechanics, but this isn’t one.

When I first encountered this absurd concept (sorta "the only real physics is that you can describe using closed forms; numerical methods are not physics") I was gobsmacked. I probably knew that Mills BBoBB takes this nonsense seriously (someone surely mentioned this waaaay back in this thread, or one of its predecessors), but had forgotten.

I guess optiongeek (and markie) would be very uncomfortable with the film "Hidden Figures"! While no doubt the screenwriter was at least somewhat, um, liberal with the facts, the physics/mathematics involved in the "Euler" scene are rock solid; here's a link which discusses this.

I do find it quite ironic that some with "left field" ideas do not realize just how much their ability to post to this forum depends on stuff they vehemently reject (ever tried to design the main chip used in your computer using only closed forms, optiongeek?).

 

[JeanTate]

JeanTate did not mention calculus.

I was the one who mentioned calculus. I'll try to respond to hecd2 without giving too much away.

JeanTate wants to take this one step at a time, so piggy-backing my own point(s) on top of his may have been a bit out of line. On the other hand, JeanTate was addressing markie, while I was responding to optiongeek.


Although I mentioned calculus, I didn't say anything about integrals or closed form. Not all calculus problems involve integrals or closed form solutions to integrals. Calculus has more to do with limits than integrals; I thought that point (notice the pun) might fit with JeanTate's question. Definite integrals are just one particular kind of limit.

Whether a definite integral has a solution in closed form is not important here. That is a point I have been trying to make and optiongeek has been failing to appreciate:



I suspect JeanTate's point is qualitative rather than quantitative, so I don't think it matters whether some integral that might somehow be related to JeanTate's point has a solution in closed form. I don't even think it matters whether coming up with such an integral is a problem first-year calculus students should be able to solve. I do think students in first-year calculus should be able to translate JeanTate's problem into a limit, but I suppose that depends on where you went to school, which instructor(s) taught the section of calculus you took, and how much attention you paid when you took the course.
Randell L Mills got his chemistry degree from Franklin and Marshall College. Apart from that one data point, which does not bode well, I am not really in any position to judge whether a pre-med/chemistry major at Franklin and Marshall College would have been taught how to set up that limit. Let me revise my claim to say only that people who have taken first-year calculus courses should know how to set up that limit.

I don’t think we should get much further ahead of ourselves and spoil Jean’s sequence of questions. Suffice it to say that not only do I agree with what you say above, but the integral that I was thinking of (which you have to arrive at if you want to determine the general solution to the problem) can easily and accurately be evaluated numerically, even if there is no closed form solution. But even arriving at the integral requires a bit of manipulation that is probably beyond first year calculus students. I do take your point about limits.

As for people not understanding that solutions need not be closed form, I remember a frustrating discussion with someone who insisted that the fact that there is no analytical solution to the general three body problem means that Newtonian mechanics is flawed (“wrong” was the word I think he used). Of course there are limits to the applicability of Newtonian mechanics, but this isn’t one.

(my hilites)

Not unexpectedly, markie has expressed no interest in even my first step, and optiongeek has been silent. So I won't be doing anything more on this.

A looong time ago, when I was getting my teaching certificate, we learned about "discovery learning". When I have used what I think of as discovery learning, for a significant number of my students, the results were spectacular: not only did they learn the concepts etc, but unlike so much learned rote, they stuck. For decades. It's true for me too: when I discover something by working on my myself, it tends to stick far better than just studying a chapter or so of a textbook.

However, a key aspect, right at the start, has to be a willingness to learn; sadly, markie and optiongeek have shown - over many, even hundreds, of posts - that they have no such interest or intent.

 

[The Man]

By orbital velocity I'm referring to tangental orbital velocity. You may be confusing it with escape velocity of something thrown straight up.

Nope, no confusion here, so how about answering the question?

 

[The Man]

Yes it is a bit of a misnomer. Only an aspect of the wave function collapses. One aspect (like position) becomes a discrete value and 'known'. The complimentary (or conjugate) aspect (momentum) then becomes the opposite of known and truly loses any kind of defined value. There goes another classical conservation law down the drain. Of course I'll side with Einstein and say hogwash.

Nope not what Einstein said, so stop pretending to be Einstein. As you have to impart or even take away some momentum to/from the particle to determine its position the resulting uncertainty of momentum in no way violates any "classical conservation law".

 

[The Man]

The loop is not deformable, it is rigid because it is kept at very high force balance tension. That tension is between inward coulombic force and outward centrifugal force. And again, the restoring force has to do with orbital mechanics of motion, not a coulombic restoring force from the shell.
You are only speculating that it is not stable.

Again, a outward force equal to the centripetal force simply means no centripetal acceleration and thus no orbit. "orbital mechanics of motion" has no such "restoring force" as you assert. You are not only merely "speculating" about orbital mechanics you are speculating in a way inconsistent with "orbital mechanics of motion" and just basic mechanics in general.

The orbitsphere under minor perturbation remains stable, while the whole atom is perturbed and moves. The orbitsphere is constrained to have a constant spin angular momentum and a constant mass, and thus, a constant velocity, and thus remains at a fixed orbital distance from the nucleus, which is another way of saying that the proton is locked into the centre.

That you or Mills simply constrain the orbitsphere so (proton is locked into the centre) in no way requires the physics and dynamics to do likewise.

 

[The Man]

Exactly the type of thinking that concludes that two photons crossing at 180 degrees each must have their velocity equal to zero for a brief instant. Unphysical and ridiculous.

If they have "velocity equal to zero" then they aren't "crossing". Do please let us know when your "type of thinking" can at least agree with, well, your "type of thinking".

 

[markie]

Nope not what Einstein said, so stop pretending to be Einstein. As you have to impart or even take away some momentum to/from the particle to determine its position the resulting uncertainty of momentum in no way violates any "classical conservation law".

Einstein wouldn't deny that the measurement process would disturb an objects' momentum. That is just a popularization of the idea and is not what QM teaches. QM teaches that an object doesn't have a definite momentum when it is not measured. Einstein would not agree to that.

 

[markie]

If they have "velocity equal to zero" then they aren't "crossing". Do please let us know when your "type of thinking" can at least agree with, well, your "type of thinking".

Hey it's not my thinking, it's Hecd's thinking.

 

[markie]

Call me stupid, but I was actually trying to help you with your communication efforts.

First, your posts have given the impression that you do not understand the key parts of classical physics (mechanics) which Mills insists is all you need, and which underlie the "orbitsphere" concept. My suggestion was for a concrete way to show that this is not so.

Second, if you think what I am suggesting something "that does not pertain to the orbitsphere", then I think this illustrates yet another of your failures of communication.

You see, I thought that an orbitsphere includes a ring (hoop, ribbon), an infinitely thin, charged, perfect circle with the charge distributed uniformly (infinitely many such rings in fact).

But OK; you know the saying about horses, water, and drinking?

Mills' rings have an infinitesimal charge and mass, unlike what you are proposing. And they are moving. And they are precessing.

 

[The Man]

Hey it's not my thinking, it's Hecd's thinking.

Hey, you making a self contradictory statement and then simply trying to ascribe it to someone else doesn't make it anything other than just your thinking.

 

[Dave Rogers]

Mills' rings have an infinitesimal charge and mass, unlike what you are proposing. And they are moving. And they are precessing.

For every infinitesmal charge and mass in an orbit, is there another equal charge and mass on the exact opposite side of the orbit, orbiting in the same direction?

Dave

 

[JeanTate]

I'll put this down to your continued inability to communicate your ideas and/or your poor reading comprehension (though it may be due to an even more basic misunderstanding of, well, the basics, than I had realized). In any case, I'm done with trying to help you.

Mills' rings have an infinitesimal charge and mass, unlike what you are proposing.

You might want to read what I wrote. Again.

Or maybe not ...

And they are moving. <snip>

Really?

Relative to what (are they moving)?

What is their velocity? acceleration?

 

[The Man]

Einstein wouldn't deny that the measurement process would disturb an objects' momentum. That is just a popularization of the idea and is not what QM teaches. QM teaches that an object doesn't have a definite momentum when it is not measured. Einstein would not agree to that.

Again, stop pretending to be Einstein. Einstein's objection was to the Copenhagen interpretation of QM. Einstein even proposed an experiment to test those objections.

https://en.wikipedia.org/wiki/EPR_paradox

So now you don't think "There goes another classical conservation law down the drain"?

 

[markie]

For every infinitesmal charge and mass in an orbit, is there another equal charge and mass on the exact opposite side of the orbit, orbiting in the same direction?

Dave

Given an infinitesimal point charge and mass on an infinitesimal ring charge and mass, yes there would be another equal charge and mass on the same ring, on the exact opposite side and orbiting in the same direction along the ring.

 

[markie]

I'll put this down to your continued inability to communicate your ideas and/or your poor reading comprehension (though it may be due to an even more basic misunderstanding of, well, the basics, than I had realized). In any case, I'm done with trying to help you.


You might want to read what I wrote. Again.

Or maybe not ...


Really?

Relative to what (are they moving)?

What is their velocity? acceleration?

You wrote, my bold :
"Consider a point charge, and a rigid, charged ring (an infinitely thin, perfect circle). Let the total charge on the ring be equal to that of the point charge. Let the charge on the ring be distributed uniformly. Assume no other forces. Assume the ring and point are not moving, relative to each other. And so on."

Again, the total charge on a ring in Mills' orbitsphere is infinitesimal, certainly not the same as that of the central point charge proton.

 

[Dave Rogers]

Given an infinitesimal point charge and mass on an infinitesimal ring charge and mass, yes there would be another equal charge and mass on the same ring, on the exact opposite side and orbiting in the same direction along the ring.

So every infinitesmal point charge has another in its L3 point. While you're explaining elementary orbital mechanics to a bunch of physicists, would you like to comment on the stability of this arrangement?

Dave

 

[markie]

Again, stop pretending to be Einstein. Einstein's objection was to the Copenhagen interpretation of QM. Einstein even proposed an experiment to test those objections.

https://en.wikipedia.org/wiki/EPR_paradox

So now you don't think "There goes another classical conservation law down the drain"?

Einstein's objection was more broad than to just the Bohr interpretation. He was opposed to indeterminacy and the statistical model arising from that. He was a believer in physical causation.

 

[markie]

So every infinitesmal point charge has another in its L3 point.

Sure, and L4 and L5 points too. Is that a problem?

While you're explaining elementary orbital mechanics to a bunch of physicists, would you like to comment on the stability of this arrangement?
Dave

Sorry, don't know enough about stability to comment much. Let's just say that with unalterable orbital velocity and with precession of a given ring into the third dimension, the picture deviates somewhat from typical perturbation concerns around orbital mechanics.