238th book of science for AP // Proving the Principle of Maximum Electricity Production is done by Atoms
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Archimedes Plutonium
Mar 18, 2023, 10:02:09 AM
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So many times I have referred to this principle in my work. Yet I never proved it true. I think it is
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Archimedes Plutonium
Mar 18, 2023, 11:21:39 PM
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Alright I need some math data to understand why Maximum Electricity production relates directly to
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Archimedes Plutonium
Mar 18, 2023, 11:42:18 PM
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Now looking at the unique features of the Torus. Source StackExchange: The torus is the only surface
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Archimedes Plutonium
Mar 19, 2023, 2:38:02 PM
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Now the speed of measured Alpha particles from decay is in the range of 5 to 7 percent the speed of
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Archimedes Plutonium
Mar 19, 2023, 11:51:08 PM
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Recently I caught myself writing a trio of books in astronomy, starting with the concept of Stepping
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Archimedes Plutonium
Mar 20, 2023, 2:27:00 AM
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Alright, I think I have the proof that the torus is the geometry figure of Maximum Electricity
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Archimedes Plutonium
Apr 23, 2023, 2:14:10 PM (6 days ago)
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Alright, onto my 238th book of science. I have often mentioned this principle in my physics work.
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Archimedes Plutonium
Apr 23, 2023, 5:06:17 PM (6 days ago)
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I suspect the Maximum Electricity Principle is the reverse of Least Action or Least Energy principle,
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Archimedes Plutonium
Apr 27, 2023, 11:27:21 PM (2 days ago)
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I need a full book on Maximum Electricity Production Principle in order to say the S, P, D, F
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Archimedes Plutonium
Apr 28, 2023, 9:24:25 AM (yesterday)
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I am going to try for 2 different methods of proof of Maximum Electricity Production. One method is
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Archimedes Plutonium
Apr 28, 2023, 12:06:03 PM (yesterday)
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On Friday, April 28, 2023 at 9:23:19 AM UTC-5, Archimedes Plutonium wrote: > I am going to try for
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Archimedes Plutonium
Apr 28, 2023, 9:32:30 PM (yesterday)
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Alright, I am not going to have any problems with figuring out the geometry inside of Atoms, for I
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Archimedes Plutonium
Apr 28, 2023, 11:20:05 PM (yesterday)
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On Friday, April 28, 2023 at 9:30:14 PM UTC-5, Archimedes Plutonium wrote: > Alright, I am not
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Archimedes Plutonium
Apr 29, 2023, 5:57:07 AM (19 hours ago)
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On Friday, April 28, 2023 at 11:19:34 PM UTC-5, Archimedes Plutonium wrote: > On Friday, April 28,
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Archimedes Plutonium
Apr 29, 2023, 4:09:32 PM (9 hours ago)
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Now this total overhaul of the geometry of the interior of all Atoms is going to make me say that the
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Archimedes Plutonium<plutonium.archimedes@gmail.com>
Apr 29, 2023, 11:57:22 PM (1 hour ago)
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to Plutonium Atom Universe
On Saturday, April 29, 2023 at 5:56:34 AM UTC-5, Archimedes Plutonium wrote:

Copper is 1s-2, 2s-2, 2p-6,3s-2,3p-6,3d-10,4s-1 Of course copper is element 29.

Now I am wondering about the mathematics here involved with nesting of toruses. I feel it is some form of Counting Principle, not a factorial of Permutation or Combination but rather : To find the number of ways of making several decisions in succession,
multiply the number of choices that can be made in each decision.

I am going to use Copper for it has only 7 orbitals. The Mainframe is the 4s-1 and holds inside its torus all the other 6 orbitals.

So the 3d-10 is inside of 4s-1.
The 3p-6 is inside of 3d-10.
The 3s-2 is inside of 3p-6.
The 2p-6 is inside of 3s-2.
The 2s-2 is inside 2p-6.
The 1s-2 is inside of 2s-2.

So the two muons of 1s-2 as they go around inside the 1s-2 proton torus manufacture electricity. Total individual bar magnets if we counted a muon as bar magnet and each individual torus as a bar magnet, is 2.

The two muons of 2s-2 plus the 1s-2 torus (three entities acting as bar magnets inside of the 2p-6 proton torus) producing electricity. Total individual bar magnets if we counted a torus as a bar magnet, is 3.

The six muons of 2p-6 plus the 2s-2 torus, plus 1s-2 torus all inside of 2p-6 and producing electricity. Total individual bar magnets if we counted a torus as a bar magnet, is 8.

The two muons of 3s-2 plus the 2p-6 torus, plus the 2s-2 torus, plus the 1s-2 torus all inside the 3s-2 torus producing electricity. Total individual bar magnets if we counted a torus as a bar magnet, is 5.

The 6 muons of 3p-6 plus the 3s-2 torus plus 2p-6 torus, plus 2s-2 torus, plus 1s-2 torus all inside 3p-6 and producing electricity. Total individual bar magnets if we counted a torus as a bar magnet, is 10.

The ten muons of 3d-10, plus the 3p-6 torus, plus 3s-2 torus, plus 2p-6 torus, plus 2s-2 torus, plus 1s-2 torus all inside 3d-10 torus producing electricity. Total individual bar magnets if we counted a torus as a bar magnet, is 15.

The one muon of 4s-1 torus, plus the 3d-10 torus, plus the 3p-6 torus, plus 3s-2 torus, plus 2p-6 torus, plus 2s-2 torus, plus 1s-2 torus all inside the 4s-1 torus producing electricity. Total individual bar magnets if we counted a torus as a bar magnet,
is 7.

So now I have two extreme cases of Atomic interior geometry that one of which is the Maximum Electricity Production via Faraday law.

The one extreme is all the protons form one big torus with the muons as a chain going around inside of 29 protons eaching having 840 windings. And here the muon chain is 29 muons.

As compared to the other extreme of nested toruses. So if we said each muon is a bar magnet and each torus going around is another bar magnet. In the case of Nested Toruses we have a summation of bar magnets as that of 2+3+8+5+10+15+7 = 50 individual bar
magnets.

Contrast 50 bar magnets in Nested toruses with that of only 29 bar magnets in one big torus.

I suppose there is a mathematical permutation or combination that can find the answer of 50, rather than doing it by hand calculation.

And this data is what I use as proving evidence that the Nested Torus is more efficient and is Maximum Electricity Production in the Faraday law.

AP, King of Science


Read my recent posts in peace and quiet. https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe   Archimedes Plutonium

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Now here I have to stop for a moment and worry and wonder of a factor in efficiency. The worry is that I know maximum electricity production is when the bar magnet is up close against the coil windings, with almost no space between them. So that if the
magnet is loose in the thrust is less production of electricity compared to when the magnet is shortened and fat rather than loose. So the magnet is almost rubbing up against the coil windings. And possible even rubbing yields more electricity.

So I wonder where in this efficiency formula is that concern addressed? Is the rubbing or almost rubbing in the B factor-- the strength of the electromagnetic field? Or possible we need a fourth factor of how close the magnetism is to the coil. And we
also need a factor of at the thrust at a angle. For we can only have efficiency of electricity production of that which is perpendicular between coil and bar magnet.


Here I think I may be bound to the concept of Efficiency of Motor with its formula of The strength of your generator depends on:

--- quoting Edison TechCenter.org on Internet ---
"L"-Length of the conductor in the magnetic field
"v"-Velocity of the conductor (speed of the rotor)
"B"-Strength of the electromagnetic field

You can do calculations using this formula: e = B x L x v

Efficiency of electric generator is e = B x L x v

End quoting EdisonTechCenter.org for their writing style is superior as educational—

Using that formula I backpedal to see if L or B is made maximum with a nested torus design or in a chain of muons design. Of course there are other possible geometry designs.

AP

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So I ask the question again is there a mathematics easy way of computing that 50 rather than have to hand walk through the calculations. I have never run across a application of mathematics Combinations or Permutations of nested and Counting Numbers in
physics before.

Say for Silver - 47, if we assume the one huge proton torus we have a chain of 47 muons inside a single one proton torus of 47 protons strung as one torus. But if we had the Nested Torus of smaller toruses nested inside one Mainframe torus of 5s-1, what
would be the number of bar magnets much higher than 47 in the chain of muons?

Does math have a easy formula that I need not hand walk through to find the answer. Or then a computer can easily be set up to calculate all the nested torus numbers.

AP

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So I am intrigued here because I am making the case that photons and neutrinos from dipoles. That a Light Wave is 1/2 photon and the other 1/2 is neutrino as a dipole of a circuit.

And then the total function of photons and neutrinoes is the wiring system inside of Atoms, composing the 840 Windings of a 840 MeV proton torus. The muon has 105 Windings and does the Faraday law with the proton inside a Atom.

It is hard to imagine in the Rutherford, Geiger, Marsden experiment of gold foil in early 1900s, hard to imagine that the ingoing alpha particle into a gold Atom interior can breach that proton torus mainframe so easily, and be shot back to the source at
180 degrees with increased velocity. Hard to imagine that the ingoing alpha particle looks at the windings and is never bothered by those windings. For a gold atom at 79 protons would be 79 x 840 windings = 66,360 Windings. Hard to imagine the ingoing
alpha particle being scattered only when it runs into a S,P,D,F orbital torus head-on collision in the opposite direction, but never bothered by the 66,360 windings.

So, what I believe solves this Windings problem, is that the Windings are made out of photons and neutrinos and so a superposition of ingoing alpha particles with Windings makes for a ease of entrance of alpha particle with interior of gold atom.

And this is probably proven true because of photon emission of atoms. In the Old Rutherford-Bohr model, photon emission was thought of the 0.5MeV particle going from one orbital to another orbital producing a photon emission. In AP's torus geometry, the
photon emission now becomes if one of the 840 windings of a proton, or possible one of the 105 windings of a muon, breaks apart and is emitted outside the atom as a dipole of 1/2 photon and 1/2 neutrino. Of course we can have monopole emissions of just a
photon or just a neutrino.

Anyway, what I am getting at here is that since the Rutherford-Bohr model is dead and fakery, we have to account for photon emission in a different way. And the best way is to make a function, job and task of the photon and neutrino-- as being monopoles
and dipoles as the Atom's wiring system.

AP, King of Science

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Alright this leads directly into the idea that the composition of the Proton as a 840MeV torus and the Muon as the true electron of Atoms at 105MeV are both composed of magnetic monopoles or dipoles of photons and neutrinos.

This draws me closer to a look at Energy = mass * (speed of light)^2. For then the only difference between mass and energy is a factor of light speed.

What this seems to say is that Atoms are standing Light Waves while electricity or Lightning bolts or star radiation of EM are dynamic Light Waves. And here the only difference I can make of this is the shape of the closed loop circuit.

EM radiation having a different shape than the EM composing the windings of protons or electrons or neutrons.

This would also change our view of the Chemical bond. Especially the metallic bond and would be that the neutrons as skin cover of proton toruses form rectangular parallel plate capacitors. So that the metallic bond can be a entire face of a rectangular
box bonded to another atom's entire face. This would allow ductile and malleable.

In Old Physics, Old Chemistry, the metallic bond was vary vague and obfuscating.

The covalent bond would be where the monopole of one atom joins with the monopole of another atom and thus a dipole as the bonding agent.

The ionic bond would be the same mechanism, and traditionally it is taught in classes that the ionic and covalent are the same.

AP

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On Sunday, April 30, 2023 at 8:16:55 PM UTC-5, Archimedes Plutonium wrote:

Alright this leads directly into the idea that the composition of the Proton as a 840MeV torus and the Muon as the true electron of Atoms at 105MeV are both composed of magnetic monopoles or dipoles of photons and neutrinos.

This draws me closer to a look at Energy = mass * (speed of light)^2. For then the only difference between mass and energy is a factor of light speed.

What this seems to say is that Atoms are standing Light Waves while electricity or Lightning bolts or star radiation of EM are dynamic Light Waves. And here the only difference I can make of this is the shape of the closed loop circuit.

EM radiation having a different shape than the EM composing the windings of protons or electrons or neutrons.

Alright, I think I know what this difference in shape is. That the EM Spectrum has the middle band region as Double Transverse Waves, while the ends of the Spectrum are Longitudinal Waves. Longitudinal waves are the motion of closed loop circuits, while
the Double Transverse waves can be either Standing or Dynamic. The Longitudinal Wave is apparently only dynamic.

This explains why in Old Physics and Old Chemistry, they had to have their high frequency EM waves-- X rays gamma rays etc visit the nucleus.

But what is strange is that the radio waves are longitudinal waves, and explains why they easily penetrate deep into buildings. In fact some buildings can amplify these longitudinal radio waves.

This would also change our view of the Chemical bond. Especially the metallic bond and would be that the neutrons as skin cover of proton toruses form rectangular parallel plate capacitors. So that the metallic bond can be a entire face of a

rectangular box bonded to another atom's entire face. This would allow ductile and malleable.

In Old Physics, Old Chemistry, the metallic bond was vary vague and obfuscating.

The covalent bond would be where the monopole of one atom joins with the monopole of another atom and thus a dipole as the bonding agent.

The ionic bond would be the same mechanism, and traditionally it is taught in classes that the ionic and covalent are the same.

AP

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Alright, this is becoming extremely interesting. I have the world divided between either a Transverse Cycloid wave or a Longitudinal Wave. Before in physics, the world was divided between particle and wave. Trouble with particle and wave duality one
cannot form the other.

With Cycloid versus Longitudinal, a ring of the longitudinal wave acts as a circle rolling on a straightline forms the cycloid wave. And with AP's cycloid versus longitudinal, I build all particles from cycloid waves if we allow cycloid to be semicircle.

Interesting, I have to see how to form a semicircle transverse wave from a cycloid structure.

A proton torus is merely its windings. A muon as electron is merely its windings, both proton and muon perpendicular to one another.

So if the world is cycloid waves, how do I form semicircle waves?

AP

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I am looking on how to recover a Semicircle wave in the manner that a cycloid wave is constructed.

The natural first exploration would be to roll a ellipse. Perhaps on a level surface or perhaps on another ellipse or even a cycloid.

This appearred in MathStackExchange showing that the ellipse rolled on identical ellipse does not form a circle (semicircle).

--- quoting MathStackExchange ---
I'm struggling to prove the following.

Set one ellipse in contact with a congruent one so that the minor axis of one is aligned with the major axis of the other. Now roll one round the other. The locus of the centre of the rolling ellipse is a circle centre the centre of the other, radius a +
b.

Is there an obvious line of attack?

(Counterexamples in replies.)
--- end quoting MathStackExchange ---

Sometimes a Google search hinders one in finding what they want. For example I am looking for how to get a circle from rolling, but Google hits are all about using the circle and getting a cycloid.

So AP wants a hit where we use a cycloid given and roll upon that cycloid figure some other figure to obtain in the end a circle (semicircle). Almost useless to Google search for their search cannot understand I seek the reverse.

This is probably the very same flaws in all robots, AI or not AI, they cannot understand the concept of reverse.

AP

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Alright, some good signs in all of this mess. I am looking for rolling of closed curves that recovers or retrieves the Semicircle. While the rolling of circle on flat surface gives the cycloid. How to recover a semicircle is what I want.

Looking at Roulette Curves, I see the rolling of a circle upon another circle the center of rolling circle etches out a Trochoid.

Now we use an off-center to the rolling circle upon another circle and we get a Limacon Roulette curve.
So almost instantly I realize if we use a point on the rolling circle we get a new circle.

This indicates to me that using a rolling circle on the cycloid with a point marker on the circumference starts to turn the cycloid into a Limacon shape. This then allows me to guess that a specific ellipse for a given cycloid curve will generate a
Semicircle when rolled upon the cycloid curve.

This is the answer I was seeking. To have a given cycloid curve, how does one restore a Semicircle from that cycloid. And the answer appears to be that you roll a specific ellipse upon the cycloid curve.

AP

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Alright, some good signs in all of this mess. I am looking for rolling of closed curves that recovers or retrieves the Semicircle. While the rolling of circle on flat surface gives the cycloid. How to recover a semicircle is what I want.

Looking at Roulette Curves, I see the rolling of a circle upon another circle the center of rolling circle etches out a Trochoid.

Now we use an off-center to the rolling circle upon another circle and we get a Limacon Roulette curve.
So almost instantly I realize if we use a point on the rolling ellipse (replace rolling circle) we get a new circle.

This indicates to me that using a rolling circle on the cycloid with a point marker on the circumference starts to turn the cycloid into a Limacon shape. This then allows me to guess that a specific ellipse for a given cycloid curve will generate a
Semicircle when rolled upon the cycloid curve.

This is the answer I was seeking. To have a given cycloid curve, how does one restore a Semicircle from that cycloid. And the answer appears to be that you roll a specific ellipse upon the cycloid curve.

AP

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Now looking at the Limacon, if we stipulate that the offcenter point has to always be as a perpendicular to the surface of the circle it is rolling over, that we generate a semicircle.

This gives me the idea that if we stipulate a ellipse rolling over a identical ellipse- it generates a semicircle.

Also, if we stipulate a cycloid rolling over a identical cycloid with offcenter marker always perpendicular to surface of cycloid rolled over, generates a semicircle.

I am surprised no-one in math or physics did this before AP does it in May of 2023, fresh new mathematics on the world scene.

AP

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On Monday, May 1, 2023 at 1:02:23 PM UTC-5, Archimedes Plutonium wrote:

I am looking on how to recover a Semicircle wave in the manner that a cycloid wave is constructed.

The natural first exploration would be to roll a ellipse. Perhaps on a level surface or perhaps on another ellipse or even a cycloid.

This appearred in MathStackExchange showing that the ellipse rolled on identical ellipse does not form a circle (semicircle).

--- quoting MathStackExchange ---
I'm struggling to prove the following.

Set one ellipse in contact with a congruent one so that the minor axis of one is aligned with the major axis of the other. Now roll one round the other. The locus of the centre of the rolling ellipse is a circle centre the centre of the other, radius a

+ b.

Is there an obvious line of attack?

(Counterexamples in replies.)
--- end quoting MathStackExchange ---

Now the title of that MathStackexchange is "Rolling ellipses" question asked 9 years, 8 months ago.

And watching the animation the one ellipse traces out almost, say 99.9% of a semicircle.

But I wonder, with my keen math intuition, that if the center point was Free to rotate and if that center point were a marker so that it could be perpendicular to the surface of the second stationary ellipse, that the two ellipses indeed, trace out a
circle.

Just wondering. And I need to make cut-outs of cycloids and ellipses and hand roll them. We cannot trust computer animations.

AP

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So, well, Wikipedia has a animation of a Limacon and the Convex Limacon is almost a circle but a part of it looks flattened.

My objection is that the point-marker is allowed to rotate. If the point marker was always staring of the circle it was revolving around at a perpendicular, then that etches out another circle.

In the MathStackExchange of a ellipse moving around the same ellipse, they almost get a circle. But I wonder if they programmed the computer animation to not be perpendicular pointer marker and thus be of the path of a circle ever so slightly.

Now it is intuitive that if the pointer-marker in Limacon was fixed stationary and perpendicular to other circle at all times that we etch out concentric circles depending on where from the center the pointer-marker is placed. As if the pointer-marker
was caught in a magnetic field that did not allow it to move from being perpendicular.

Now if the pointer-marker were directly on the rim, the circumference starting out, can we expect the result to be a circle, only a circle with one point intersection with the circle it is rotating around in 360 degrees.

These figures are actually far more complicated and complex than what Wikipedia and Websites make them out to be.

AP

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Now I am going to go contrary to the description of the Dimpled Limacon and say that the Pointer-Marker needs to stay Perpendicular to the tangent of the surface of stationary circle. In this manner, the dimple and the flattening disappears and what we
end up with is a perfect circle, larger than stationary circle and intesection of both circles at one point.

AP

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Now I am pretty sure this is a flaw in Old Math Geometry of their Limacon pointer marker going around like a clock as the motion of the moving circle goes around. The AP Limacon is where the pointer marker needs to stay perpendicular to the tangent of
the stationary circle, thus creating a new larger circle with one intersection point. I say this because there is no other mechanical means of producing concentric circles or intersection at one point of two circles.

Physics needs the **perpendicular** for the pointer-marker, for electricity is always perpendicular to magnetism.

The Old Math Geometry Limacon is actually 2 rollings, the rolling of one circle over another and the independent rolling of a pointer marker. In the AP Limacon, there is but one rolling the circle over the other circle, while the pointer-marker is a
fixed perpendicular point.

This messy obfuscation of what is moving also can solve the two identical ellipses rolling one over the other as tracing out a perfect circle. It is almost producing a perfect circle in the MathStackExchange animation but slightly off. And I am saying it
is only slightly off because the animation was not programmed to allow the pointer marker, the center of the rolling ellipse to always have a perpendicular to the tangent of the stationary ellipse. You see the programmers of that animation, are like the
Limacon animation and not restricting the pointer-marker to be always a perpendicular, and that is the reason it is not a perfect circle.

Now I need to do the same thing with the cycloid, see if I can roll one cycloid over a stationary cycloid with a pointer-marker and see if I trace out a semicircle. I may have to use the midpoint of the line segment that connects the two ends of one
cycloid wavelength.

Yes, I believe that returns two cycloids, one rolling on the other, to be a semicircle.

AP, King of Science

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