The Cause of Gravity - the next major chapter

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Nevyn wrote. Something I think we have all failed to remember, is that Miles has switched the vectors around. He is proposing that gravity is not the 9.81 down and earth's emission is not the 0.009545 up. He has switched them around so that the earth's emission is 9.81 up and the gravity vector is 0.009545 down. Although I probably shouldn't call it a gravity vector.
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That is the mechanism that Miles is using to explain how bodies don't actually feel the full 9.81 up vector. Most of the charge is not hitting you, it is being channeled through you.

Now, how exactly that happens is still a mystery to me.
Airman. Hey Nevyn, I won't pretend to understand the 9.81m/s^2 up (emission resulting in charge repulsion and charge binding) and the 0.009545m/s^2 down (my mind is a blank). I believe what is most important about molecular binding is the object's ability to turn the upward moving emission to the sides. Stating that more clearly, matter turns the received charge orthogonal to the emission field, which provides the source charge for binding molecular matter. That's related to lift, and the orientation of a nucleus, coherent with the earth charge emissions.

With respect to your moon comments, I had every intention of giving them some thought before attempting an answer earlier today; Vexman beat me to it.

Vexman, for the record, those are great comments in your last post. I have no disagreement with any of them. I was waiting to see if you might get any other responses, but the field is sparse. If it was up to me, I'd declare you the winner of this discussion.
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Vexman wrote:

Nevyn wrote:To me, the idea of charge alignment is a big problem. We can't even see how it would align to a single body, so how is it going to align to multiple bodies? Should the Moon align to the Earth or to the Sun? The obvious answer is to the earth, because that is what it is orbiting. However, the moon doesn't exist only in the earths charge field. It is at all times also in the sun's charge field. Yet the moon is in gravity-lock with the earth, so it is acting like the sun has no effect on it. The moon is 60 earth radii from the earth, well outside the 11 earth radii of the charge-pause, so the sun's charge field should dominate. The moon isn't even in the earth's charge field. It is no where near it.

That's not entirely true. We may not be able to see the incoming photon charge field aligning to a particular larger body / mass, but we certainly can see a larger mass aligning to the dominant charge field (as a source). For instance, that's the reason all planets orbit around the Sun in its equatorial plane, rather than orbiting in all possible planes around the Sun. So the alignment is happening and we do understand it to a point. I don't see any real argument here which would negate the idea of incoming charge alignment per se as impossible or not mechanically feasible.

I was referring to the alignment of the nuclei of planets or moon or any body in a gravitational field, not the charge itself. Sorry, bad wording to use. The point is that it can only align to one body, not multiple. The alignment seems necessary, to me, since it is using through-charge of the gravitated body to cause spin-ups to cause the gravity vector. Through-charge needs to go straight through the nucleus, without collision with that nucleus. So the nucleus must have its own N/S axis in the same line as the emission from the source of gravity. How can the moon align to both the earth and the sun? The problem is the need for an alignment. That alignment excludes other bodies from exerting a gravitational influence, except when they are actually in alignment themselves and the nuclei are between them, but that is a fleeting moment.

Vexman wrote:You should as well include density of the charge field as it has a large impact on the local charge field  order. In the case of Moon-Earth-Sun while looking at the Moon: the Earth has much denser charge field compared to the Sun's simply due to its vicinity to the Moon, even though it is much smaller in mass. The Sun's emitted charge photons are also being hit throughout their journey, diminishing such field's coherence (of linear direction and/ or spin energies) for longer amount of time if compared to the Earth's. Considering the distances which have crucial impact on charge field's density , Earth's charge field toward the Moon has more density and thus carries more binding force. I believe this is the reason we see the Moon locked-in to the Earth, as it has enough needed force to overcome Sun's force, doesn't it? What we can observe is then the balance of these forces.

So local density of the charge field plays a crucial role. The basic characteristic of the charge field is that it is not homogeneous, it's chaotic in sense of movement direction and spin energies (therefore sizes). Larger mass starts to change this field's characteristic as soon as it starts recycling charge. So by the order of appearance - the first charge body doesn't align to anything as there is no dominant charge source. But the second charge body will align to the first and vice versa, if they are close enough, depending on the "power" of recycling engine - the bigger the mass, the more dominant and far-reaching is its charge field.

I did include density, just not in so many words. I discussed the distance of the moon in relation to the distance of the charge-pause. I also stated that I think the earth's charge field does reach the moon (precisely because of orbits) but it can't be the dominant field at that distance. It only has to reach the distance to provide exclusion, and that is all I needed for an orbit. It is binding gravity that requires alignment, and hence it requires that the earth be the dominant field at the moon. Yet we know that it isn't, because of the charge-pause.

Vexman wrote:On top of all that, I don't agree with the idea of a charge pause at all. The charge field is apparently omnipresent so there can be no charge pause in physical terms where it (the pause) means there is no charge present. Maybe at the 11th Earth radii, the Earth's emitted charge loses its ability to create and sustain the E/M field, creating the so-called magnetopause at that distance. But it doesn't imply or say anything about the end-reach of emitted charge photons not being able to cross that imaginary line. What would make photons abruptly stop en massè, anyway?

You are misunderstanding what the charge-pause is. It is not the end-of-the-line for the earths emission. It is not the point where charge stops, or pauses. It is the point where the earth's charge field and the sun's charge field are of equal strength, which actually means equal density. Therefore, a body at that point would not feel any net force towards one or the other, ignoring other forces. It is being pushed equally from both sides. The earth's emission keeps on going and so does the sun's, but the densities are different. Outside of that sphere, the sun's field is more dense, but inside of it, the earth's is. Outside, the sun is dominant. Inside, the earth is dominant.

Vexman wrote:I also don't agree that Earth's charge doesn't reach Moon. Sure it does, but it's not as dense as it would have been at a closer distance. If we compare it to the density of Sun's emitted charge at Moon's surface, I would think that Earth's emitted charge is still more dense than Sun's. Earth's charge has relatively larger impact than Sun's but only locally. That is why the Moon is locked to the Earth, the local charge field is dominated by the Earth rather than the Sun.

I agree that the earth's charge can reach the moon, and stated so, but I don't agree that the earth's charge field can do so. There is a difference, all-be-it a technical one. In order to call it a field, it has to dominate. If some other field is more dominant at some point, then you are in the field of that dominant source. Of course there is some overlap. I would say that we can still call the earth's emission a field for a little way past the charge-pause, because it still has some appreciable density and structure. It isn't dominant, but it still has an appreciable effect. Can we take that all the way out to the moon though? I don't think so. It is over 5 times the distance of the charge-pause, where we know that the sun is equal. Maybe we could allow another 11 earth radii, just to be generous, to that and say that the earth's charge field extends to 22 earth radii from the surface. That is still only a third of the way to the moon, and density drops off with an inverse quad relationship.

Vexman wrote:

Nevyn wrote:My main point was that it can only be aligned to one source at a time. If gravity requires that alignment, then a single body can only feel gravity from a single source. That rules out Lagrange Points because they rely on the intersection of multiple gravitational fields.

Alignment is only a general fact, it doesn't mean that absolutely all nuclei of a larger mass need to align to the same source simultaneously. Alignment is also a process, not a single event or occurence. Looking at the larger bodies, for instance the Moon, it aligns most of its nuclei to the Earth for the most part, but it also aligns smaller portion of them to the Sun, simultaneously. Moon's position in the orbit around the Earth is then in perfect balance between the two binding forces (and repelling charge emitted by both the Sun and Earth). I don't see it necessarily as black&white, there is no particular reason for this to be so. If distances between bodies change, so does the alignment and the balance between the charge sources / forces is re-established. In essence, Lagrangian points represent the points in space where we can find all gravities (and other forces) balanced out.

I don't think we can assume that some nuclei align to one and some align to another, as such forces would rip the body apart. Molecular bonding requires that the nuclei align to each other and if that alignment is broken, then the molecule is destroyed. We are talking about solid structures. Therefore, they have bonds in all 6 directions from the core of each atom (excluding the edge nuclei). Not all molecules are bound to each other in this way, though, so it looks like we could align some to a different source, but that would rip the body apart. You need them bound to each other to stop it doing that, but you need them unbound to align to different sources of gravity. A contradiction.

You also need a charge field capable of aligning the nuclei. Only one field will be dominant for the moon, and it isn't the earth's. Alignment requires charge coming from a certain direction, and it must be the main source of charge as it must overcome all other sources. If the earth can only overpower the sun within 11 earth radii, then it isn't going to overpower it at 60. Especially when the moon is between the earth and sun. In that position, you have the moon mostly aligning to the sun and some of it aligning to the earth, but they are on opposite sides of the moon. To accommodate that, you need the moon to block the charge of the sun so that the earth's charge has more strength, but gravity has never been shown to be blockable.

This is another problem for binding gravity. Miles himself has used the non-blocking nature of gravity to explain some things, such as Coulomb's charge experiments. It is obvious to see how expansion can not be blocked, but it is easy to see how charge based gravity can be blocked. The magneto-tail shows us that directly. The earth is blocking the sun's charge, so the magneto-pause extends further away from the earth in the direction directly away from the sun. So we should see some serious changes in gravity in such locations, but we don't.

Hello everyone, I don’t post since forever. Welcome to all the new people!

I want to add my 2 cents here too because the new gravity papers were really mind blowing.

In my opinion the most important aspect of this new theory is that we go beyond simple pool ball mechanics.

Earth emission up is not only a bombardment that would push, it is actually a components of 2 types of photons. A component can be channeled and the other cannot. Only the latter causes a push.

The same applies to the charge field the Earth is in, part of it can be channeled and part cannot. Of course if a body is not spherical all of that charge cannot be channeled so it becomes a push.

In my opinion 9.81 is both the earth emission and the Earth local ambient field. The resulting 9.80 is the ratio between the recycled flux up and the bombarding flux down. But this is only my thought, I still haven’t polished this enough to explain it thoroughly, let alone proving it to be correct.

Nevyn, I can see where and why we have different opinions. The first issue is about the charge-pause and its meaning. While you support the idea of the charge pause as the limit of body's charge field, I don't think it is a correct assumption.

So we should look at the meaning of a charge pause. You describe it as:
"the point where the earth's charge field and the sun's charge field are of equal strength, which actually means equal density. Therefore, a body at that point would not feel any net force towards one or the other, ignoring other forces.

That assumption is based on the observations of the magnetosphere and its limit (aka magnetopause). What is actually observed is the boundary between the planet’s magnetic field and the solar wind (from Wikipedia). When solar wind meets Earth's E/M field, the direction of wind's flow is changed so that most of the solar wind plasma is deflected to either side of the magnetopause. I'm just paraphrasing mainstream here, trying to understand how you've come to a conclusion about charge pause. And to be honest, I don't see the connection here. The limit at 11th Earth radii is the magnetopause limit, but what does this say about Earth's or Sun's charge field, their limit or even strengths?

What I believe it means is that the solar wind would have reached Earth's atmosphere and surface if it wasn't for the E/M field (or shield in this case) deflecting it. That's what is happening to Mars or Venus with zero measured E/M presence. But both Mars and Venus still recycle charge and still emit charge. By your suggestion, the reach of their charge field is equal to 0 radii. So their binding force/gravity should be equal 0, since their charge field cannot be dominant at any given moment. As well, their orbits would be non-existing as a consequence. None of that can be observed as true, so it implies that their charge fields have a much larger field of influence. It also means that the influence zone of E/M field isn't equal to the influence zone of body's overall charge field.

That's how I understood the meaning of magnetopause, while I still think charge pause is not the correct term to properly describe it. In my opinion, the Earth's charge pause (or better, charge field zone of influence's limit) is showing in the Moon's orbit - considering the balance of all participating masses, densities, binding forces, charge fields' influence zones and E/M fields, the Moon is in the "sweet spot" where no particular force can overcome any other force. The very same can be said about any planet or satelite in our system, their orbits show the region where all forces are balanced out.

You also say that you "don't think we can assume that some nuclei align to one and some align to another, as such forces would rip the body apart. Molecular bonding requires that the nuclei align to each other and if that alignment is broken, then the molecule is destroyed."

I should have been more precise in my previous post. In my opinion it's even easier to explain this using molecules and their bonding instead of single nuclei. Important to note is the alignment being a continuous process. Even a molecule has its N - S polar vortice, which is aligned to the dominant charge source. While it could be aligned to one source for most of the time, it could still accept other source's charge - with the binding force vector pointing at different angle, the molecule would start / tend to align itself to the alternate source. But this doesn't happen in the end as the dominant source is repeatedly taking over the alignment (the density of dominant force field being larger than alternate source field's density). All of this is going on in ultra-small fractions of a second, and it's a continuous process, just like constant adjusting of the orbits while balancing out the forces. So we're not braking the molecular bonding, we're bombarding the already aligned molecule (taken over by the dominant charge field) from another angle with alternate charge field particles, where such molecule is alternating its intake between one and the other source. Photons don't come labeled or tagged by its source, the molecule can't differentiate between i.e. Earth's or Sun's photons as they are all the same. Molecule can only recycle charge that is headed towards its vortex and surface. Molecule will also align its vortice to the dominant charge source. So alternating between charge sources doesn't necessarily imply breaking the molecular bonding, but implies the tendency of a molecule to align itself (on N-S axis) toward another angle of incoming charge. We experience this as binding force coming from two sources, while in reality it is a constant "dance" of forces at play. While most molecules / nuclei will be aligned to the dominant charge field, some of them will be accepting the charge from an alternate field, while tending to align to its incoming angle. No bonding is being broken while we can still take note of two separate vectors of binding force in such process.

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I see Vexman has replied, good. I'll read it after my post.

Hey Nevyn, as I see it, given a field of many celestial bodies, there is no reason to believe that all the matter of one body would train itself to only one other celestial body. The sun is engaging with each planet all the time and each planet returns charge pretty much along those same channels.

Further, I would say that all nearby celestial bodies establish two-way channels with each other all the time, but they are generally brief and tenuous and secondary in nature. They do leave traces of charge in their orbital wakes. Normally the charge received from the bodies are cyclical events that do not upset the existing balance of the solar system.

The earth is ‘aligned’ to the sun, yet the earth is also aligned to the moon, without necessarily tearing itself apart. Although there are very significant forces at play. We can actually observe their effects. Most of the time the channels of the four main planets - Jupiter, Saturn, Uranus and Neptune appear as separate images on different sides the sun’s surface. Miles has described how the alignments of the main four planets cause of the solar cycle. I expect that the positions of the big four also help explain the earth’s seismic activity. You may recall, a surprise finding of that paper was the relative strength of Neptune’s returning charge.

True, the moon is moving through space that is dominated by solar charge emissions. The charge-pause indicates where the earth’s and sun’s charge emissions balance - the equal field virtual ‘boundary’. That field definition is one sided and fails to take the strong returning moon to earth charge into account. That two-way acceleration between the earth and moon enables the almost incredible earth/moon distance. The imbalance occurs because the moon is moving too quickly to establish a more significant return charge channel to the sun.

Ciaolo wrote. In my opinion 9.81 is both the earth emission and the Earth local ambient field. The resulting 9.80 is the ratio between the recycled flux up and the bombarding flux down.
Hi Ciaolo, thanks for chiming in. Your comment sounds like what I’m starting to think, that downward component must be returning charge.
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Vexman wrote:Nevyn, I can see where and why we have different opinions. The first issue is about the charge-pause and its meaning. While you support the idea of the charge pause as the limit of body's charge field, I don't think it is a correct assumption.

So we should look at the meaning of a charge pause. You describe it as:
"the point where the earth's charge field and the sun's charge field are of equal strength, which actually means equal density. Therefore, a body at that point would not feel any net force towards one or the other, ignoring other forces.

That assumption is based on the observations of the magnetosphere and its limit (aka magnetopause). What is actually observed is the boundary between the planet’s magnetic field and the solar wind (from Wikipedia). When solar wind meets Earth's E/M field, the direction of wind's flow is changed so that most of the solar wind plasma is deflected to either side of the magnetopause. I'm just paraphrasing mainstream here, trying to understand how you've come to a conclusion about charge pause. And to be honest, I don't see the connection here. The limit at 11th Earth radii is the magnetopause limit, but what does this say about Earth's or Sun's charge field, their limit or even strengths?

I'll let Miles answer that one:

Miles Mathis wrote:This is why Venus can also exclude the Solar Wind, even without a magnetosphere. The ability to exclude has little to do with the magnetic field and more to do with the charge field. Venus has enough mass to create a charge field strong enough to exclude the charge field of the Sun at that distance. With the Earth, the same is true. It is not the magnetosphere that excludes the Solar Wind, it is the charge field. The charge field creates both the magnetosphere and the boundary, and the boundary is only at the edge of the magnetosphere because both the charge field and the magnetosphere pause at the same place, for the same reasons. The magnetopause is at the charge-pause, because the magnetic field is caused by the charge field. The magnetic field would not be expected to extend beyond the charge field, because the charge field creates the magnetic field. And the magnetic field would not be expected to stop short of the charge-pause, because there is no reason for it to stop short. Ions are driven up by the charge field, and will persist as long as the charge field persists.

Vexman wrote:What I believe it means is that the solar wind would have reached Earth's atmosphere and surface if it wasn't for the E/M field (or shield in this case) deflecting it. That's what is happening to Mars or Venus with zero measured E/M presence. But both Mars and Venus still recycle charge and still emit charge. By your suggestion, the reach of their charge field is equal to 0 radii. So their binding force/gravity should be equal 0, since their charge field cannot be dominant at any given moment. As well, their orbits would be non-existing as a consequence. None of that can be observed as true, so it implies that their charge fields have a much larger field of influence. It also means that the influence zone of E/M field isn't equal to the influence zone of body's overall charge field.

You are treating the E/M field as the charge field, but they are not equivalent. The charge field drives the E/M field. We can have a charge field without ions present and it will still cause exclusion. So your suggestion that my suggestion is that the charge field is non-existent is false. That last sentence is wrong too. The influence zone of E/M is totally reliant on the underlying charge field. Take away the charge field and the E/M field disappears. Take away the ions and the E/M field disappears, but the charge field remains. The E/M field can be measured as being less than the underlying charge field, but it can never be more than it, and the difference is only caused by the presence of ions. The forces are still present, but there is nothing for them to act on.

Vexman wrote:That's how I understood the meaning of magnetopause, while I still think charge pause is not the correct term to properly describe it. In my opinion, the Earth's charge pause (or better, charge field zone of influence's limit) is showing in the Moon's orbit - considering the balance of all participating masses, densities, binding forces, charge fields' influence zones and E/M fields, the Moon is in the "sweet spot" where no particular force can overcome any other force. The very same can be said about any planet or satelite in our system, their orbits show the region where all forces are balanced out.

Orbits don't need the full extent of the charge field. They only require exclusion. They don't care one bit about the magnetic component. Orbits are also not using the charge field of only one body, they use them from both bodies, or all bodies involved. So there is a two-way street for an orbit, but we are discussing gravity, a one-way street. Yes, gravity is a part of an orbit, but we aren't trying to explain orbits, only the gravity component.

I may have been a bit too strict in my definition of charge field. I did limit it to the charge-pause, but an orbit does need the electrical component at a much larger distance. Which I did address. The problem is how far can we take that? At what point do you think that we can no longer call it the earth's charge field? It is totally possible that some photons emitted by the earth make it out of the solar system. Let's imagine that they reach another galaxy. Do we now say that the earth's charge field reaches that far? There has to be some practical limit as to what we call the earth's charge field and what has become just random charge photons. I suggest that that limit is where one component or the other is lost. The charge-pause is where the magnetic component is lost. The individual photons have not necessarily lost their spins, but they are now mixed with other charge that has different spins, so the magnetic field drops to 0, or whichever is the stronger field. The electrical component can continue on and still be effective, but can we still call it the earth's charge field? Maybe we should call that the earth's electrical field, since the magnetic component has been overpowered.

You are suggesting that the moon sets the limit for the earth's charge field, but it doesn't. That is the point that the earth-moon system matches in forces. A smaller body would allow the earth's charge field to be considered closer than the moon is and a larger body would make it further away (it is more involved than that and relies on density, but the point is the same). See Miles paper on Lagrange points for more on this.

Vexman wrote:You also say that you "don't think we can assume that some nuclei align to one and some align to another, as such forces would rip the body apart. Molecular bonding requires that the nuclei align to each other and if that alignment is broken, then the molecule is destroyed."

I should have been more precise in my previous post. In my opinion it's even easier to explain this using molecules and their bonding instead of single nuclei. Important to note is the alignment being a continuous process. Even a molecule has its N - S polar vortice, which is aligned to the dominant charge source. While it could be aligned to one source for most of the time, it could still accept other source's charge - with the binding force vector pointing at different angle, the molecule would start / tend to align itself to the alternate source. But this doesn't happen in the end as the dominant source is repeatedly taking over the alignment (the density of dominant force field being larger than alternate source field's density). All of this is going on in ultra-small fractions of a second, and it's a continuous process, just like constant adjusting of the orbits while balancing out the forces. So we're not braking the molecular bonding, we're bombarding the already aligned molecule (taken over by the dominant charge field) from another angle with alternate charge field particles, where such molecule is alternating its intake between one and the other source. Photons don't come labeled or tagged by its source, the molecule can't differentiate between i.e. Earth's or Sun's photons as they are all the same. Molecule can only recycle charge that is headed towards its vortex and surface. Molecule will also align its vortice to the dominant charge source. So alternating between charge sources doesn't necessarily imply breaking the molecular bonding, but implies the tendency of a molecule to align itself (on N-S axis) toward another angle of incoming charge. We experience this as binding force coming from two sources, while in reality it is a constant "dance" of forces at play. While most molecules / nuclei will be aligned to the dominant charge field, some of them will be accepting the charge from an alternate field, while tending to align to its incoming angle. No bonding is being broken while we can still take note of two separate vectors of binding force in such process.

You are missing my point. It isn't the ability of an atom or molecule to align that is the problem. It isn't that they can change alignment based on other fields. I agree that they can do that. The problem is that binding gravity requires one alignment and one only. If the nuclei are not directly in-line with the emission field of the central body, they can not create the downward gravity vector. That is because Miles is using through-charge to create spin-ups and then, after those spun-up photons leave the nuclei, they then collide with other nuclei to create the downward vector. It is pretty obvious that a downward vector requires charge photons to be moving in that downward direction to create it. Therefore, the nuclei must be aligned to the central body. Directly aligned, with no compromises.

So in the presence of multiple charge fields, how can that possibly operate? The nuclei can not align to both, or if they do, then it will mean that they align to the combination of those fields, so somewhere in between their direct alignments. That breaks the gravity vector for both bodies. That would mean that all we need to do to create an anti-gravity device is to create an electrical field strong enough to break that alignment. That doesn't seem that complicated to me. Surely it would have been found by now. It doesn't even have to change the alignment by much. We don't need a 90° change, a few degrees would do.

Vexman wrote:Photons don't come labeled or tagged by its source, the molecule can't differentiate between i.e. Earth's or Sun's photons as they are all the same.

They can and do differentiate them, based on their vectors. They don't care that it came from the sun or the earth, but they care about what direction they are coming from and possibly what spins they have. Although, care is too strong a word, it isn't sentient, it just reacts to the vectors in the moment of collision.

I just don't see how you can have a single body with its atoms and molecules being pushed in different directions, but it doesn't break apart. It has to be bound to itself to not break apart, but you have them pointing in different directions. How can you possibly think that the molecules are still bound to each other while being re-aligned to bind to different things? You are relying on fuzziness. The idea that some align here and some align there and it all equates to a single force vector is fuzzy. It isn't mechanical. For the molecules to bond to each other, they must point their charge channels at each others charge channels in order to share charge and make a larger channel. For the molecules to create the gravity vector they must be aligned to the gravitational body. You can have one or the other, but you can't have both at the same time when there are multiple sources to align to.

Vexman wrote:I'm just paraphrasing mainstream here, trying to understand how you've come to a conclusion about charge pause. And to be honest, I don't see the connection here. The limit at 11th Earth radii is the magnetopause limit, but what does this say about Earth's or Sun's charge field, their limit or even strengths?

The charge-pause gives us a measurable distance of the earth's charge field, with respect to the sun's charge field. Within that radius, the earth's charge field is stronger than all others. The earth sets the field and the rest have to fit within that. Just like the sun sets the field for most things in the solar system. By 'set the field', I mean that it is the first field, it is the primary field. It is the field that you start an equation with, and then apply adjustments to it from other sources. It takes precedence over all other fields.

Binding gravity requires 2 types of charge: inside charge that is moving out; and outside charge that is moving in. Don't think about photons and anti-photons, just reduce it to locations and velocities. It also requires something to channel both of the charges and create spin-ups on one of them, the outside charge. I assume that the something to channel is also the body that is being gravitationally attracted to the body that is emitting the inside charge. Note that it is not related to the outside charge. It is not the source of that charge. For the earth, the sun is the source. For the sun, well, I don't know. The galactic core seems too directional to be considered the source for this purpose, but I might be swayed with a good argument. The sun also seems too directional, but I'll allow a random source of charge, from all directions, for the sake of argument.

So now we are reducing it to locations, velocities and mass. We gained the mass variable from the spin-ups. The outside charge encounters the satellite. Some of it moves through the satellites nuclei as through-charge, collides with some inside charge moving in the opposite direction, and causes the outside charge to gain a spin. The outside charge is now larger than the inside charge. But it hasn't even left the satellite yet, and a lot of it won't, at least not on its original trajectory. After it leaves the nuclei that spun it up, it collides with another nuclei of the satellite. Since it has more mass, it creates a greater force than it would have before it was spun-up. This creates a larger vector down than the inside charge can create up.

That is how I see binding gravity at the moment. It may be wrong. It may be over-simplified. I don't know. Feel free to provide a better description.

Let's analyze the situation. The first thing to notice is the need for specific directions. The inside charge has a definite direction because it is being emitted by the central body. The outside charge is severely limited in its choices though. It must be coming down in a direction that allows it to collide with the inside charge. That means it has to match that charge in direction, but opposite. Not such a random source of charge anymore, is it? I believe that Miles has addressed this issue, but I can't remember the details. He discusses the day/night difference and accommodates for it somehow. I won't pick on it without further study. I will allow a not-so-random source of charge that just happens to match the emission. We don't need every charge photon to take part in gravity, so we can assume there is enough to do so. One might even suggest that it is the inside field that channels the outside field, compressing it and giving it direction, but one should be very, very careful with that.

So that brings us to the nuclei. The cause of the force differential and the reception of that force, although not at the same time for a given pair of charge photons. The nuclei are where it all happens. They are the bar, and the photons are the drunkards, falling into each other. The first thing the nuclei do is provide a nice safe place to collide in some way that causes the outside charge to gain a spin. How does it do that? By not getting in the way! A good bar let's the drunkards fall and fight, without smashing up the place. So all the nuclei are doing is providing a straight path for the inside and outside charge. That means that the nuclei must be aligned with the direction of those charge photons. It absolutely requires that alignment.

The second thing the nuclei do is receive the force created by the spun-up photon. That's pretty easy. It just needs to get hit by it. No problems there.

Given all of that, we now must ask how the nuclei can be aligned like that? What puts them into that alignment? What forces do we have to do the job? We have the emission field of the body, the inside charge. It is already in the right direction. Can we use the outside charge? It is supposed to be a some-what random source of charge, so we shouldn't be able to use it like this. Alignment requires a directional field. While I might allow some of that charge to be on the right trajectories, I can't allow most of it to be.

So all we have is the inside charge. The emission of the central body. Great, we think, for it is just perfect for the job. Is it, though?

It might be until you start to think about density and field strength. It is probably better to use strength, because that encapsulates mass (number of spins on the photons) as well as density (the number of photons per unit volume). The nuclei must align to the stronger field. That is what makes it the stronger field, from a measurement perspective. It is how we know that there is a stronger field. So if the inside charge is setting the alignment, then the central body is the stronger field.

Now, given that scenario, as I have written it, there is no problem. Well, there are some problems still, but not in the way I have been trying to get at in my latest posts. It is obvious that the body is setting the field and that the nuclei are going to align to it. There is no other field to work with. There is the outside charge field, but that is a random field so it doesn't provide the direction for alignment, as a field. That all changes when we remember that we have orbits within orbits.

We need to bring in a moon. The moon has to orbit a planet, that is orbiting a star. We aren't really looking at the star, but it is the star that creates all of the problems. The planet doesn't have empty space to emit into. It now has to emit into an existing field. The planet doesn't set the field anymore, except for in its local environment. The charge-pause is what defines that local environment.

That, my dear friends, is why the charge-pause is a serious problem to my understanding of binding gravity. Our moon is much farther away than our charge-pause. So there is no way to make the moon align to the earth's field, when the sun is providing a much larger field. Where are the outside photons coming from? The other planets, you say? Do we see a great increase in the earth's gravity applied to the moon when it is inline with the sun? Surely it has more photons on the right trajectory to create gravity in that position. It has the full force of the sun behind it. It even removes the alignment problem. I would expect a pretty big difference.

I wanted to retract my post within minutes. There was Miles' paper http://milesmathis.com/pause.html dealing with it, but I didn't want to let go of the idea and got carried away. Apparently.

Will have to re-think the solution or face the fact, that it may not be feasible. Which means I have to admit I have been defeated in this discussion so far Thank you all for your patience and creative debate so far!

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Nevyn wrote. That, my dear friends, is why the charge-pause is a serious problem to my understanding of binding gravity. Our moon is much farther away than our charge-pause. So there is no way to make the moon align to the earth's field, when the sun is providing a much larger field. Where are the outside photons coming from? The other planets, you say? Do we see a great increase in the earth's gravity applied to the moon when it is inline with the sun? Surely it has more photons on the right trajectory to create gravity in that position. It has the full force of the sun behind it. It even removes the alignment problem. I would expect a pretty big difference.
 
Vexman wrote. I wanted to retract my post within minutes. There was Miles' paper http://milesmathis.com/pause.html dealing with it, but I didn't want to let go of the idea and got carried away. Apparently.

Will have to re-think the solution or face the fact, that it may not be feasible. Which means I have to admit I have been defeated in this discussion so far    Thank you all for your patience and creative debate so far!

Airman. It’s been a great discussion between Nevyn and Vexman. The “serious problem” concerning the charge pause certainly casts doubt about Miles’ charge binding theory, so much so, Vexman appears to have yielded the field. I for one am getting a good charge field review.

The Moon and the Magnetotail
https://www.nasa.gov/topics/moonmars/features/magnetotail_080416.html


Nevyn wrote. Our moon is much farther away than our charge-pause.
Airman. As you can see from the attached image, that statement isn’t entirely true; on average, yes, as the caption says - the moon spends about six days a month in earth’s magnetotail. Those six days aren’t simply inside or outside the earth’s ion exclusion zone; the moon penetrates the magnetosphere “walls” twice each month. The moon can receive a great deal of protonic matter originating from both the earth or sun in each passage through the earth’s magnetosphere.  

Nevyn wrote. So there is no way to make the moon align to the earth's field, when the sun is providing a much larger field.
Airman. Small change - there is no way for the moon to align to the sun’s field any better, even though the sun is providing a much larger field.

The celestial sphere – including the sun - spins about the moon roughly every 24 hours. That's sufficient for charge binding as is it is true between the spinning earth and sun. A big difference is the case of the earth and moon is the fact that Luna always presents the same face to the Earth. The earth sits, spinning in the sky once every 24 hours, but it never changes its spatial location with respect to the moon.

The moon is in a molecular-like coherent alignment with the earth (aside from the earth’s rotation) which has increased in strength over time. Miles has pointed to the blasted face of the moon as due to charge from the earth. The flattening enables a broader charge binding linkage between the earth and the moon’s core, no doubt allowing a greater earth/moon orbital distance.

The sun’s field is far more extensive than the earth’s field, but at the earth/moon – sun radius, the sun’s emission field doesn’t get much stronger than the earth’s. Even if it did – to some degree, the sun is in constant orbital motion and so the sun can never establish the net channel/repulsion ratio with the moon, as the moon shares with the earth.  

Nevyn wrote. Where are the outside photons coming from? The other planets, you say?
Airman. With respect to the moon, I believe the charge received from the sun is at a fairly constant (+/- the orbital radius) rate and magnitude. On average, the Earth’s field density equals the sun’s, but the moon experiences drastic changes in the earth’s charge field density when going through the earth’s tail.

My general observation, the moon receives charge from both the sun and earth. It seems that some of the charge received from the sun – ok, or from the other planets too, will be diverted to the moon’s earth link. It seems to help explain the huge orbital radius. How does that affect gravity?

Nevyn wrote. Do we see a great increase in the earth's gravity applied to the moon when it is inline with the sun? Surely it has more photons on the right trajectory to create gravity in that position. It has the full force of the sun behind it. It even removes the alignment problem. I would expect a pretty big difference.
Airman. We do see a change in earth’s gravity during eclipses as shown by pendulum tests; I don’t agree that we should expect a pretty big difference. The coherency between celestial bodies never varies that greatly. I don’t believe you can add or subtract gravity as the planets pass in-line with each other as you seem to be suggesting. The moon should have been turned by the earth’s magnetotail wall during some grand maximum conjunction and lost long before now if it did.

I don't see the earth's magnetosphere charge-pause as a problem for charge binding.  
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You made me laugh out loud, Airman. That was not my farewell, my friend. Just made a step backward so I could see where I was wrong in my assumptions.

I really liked your point about the BPhotons and the need to rethink some essential characteristics of their "construction", just didn't have the opportunity to comment it properly. But it has a lot of reason to its arguments, which are in my opinion very hard to ignore.

Nevyn wrote:
I just don't see how you can have a single body with its atoms and molecules being pushed in different directions, but it doesn't break apart. It has to be bound to itself to not break apart, but you have them pointing in different directions. How can you possibly think that the molecules are still bound to each other while being re-aligned to bind to different things? You are relying on fuzziness. The idea that some align here and some align there and it all equates to a single force vector is fuzzy. It isn't mechanical. For the molecules to bond to each other, they must point their charge channels at each others charge channels in order to share charge and make a larger channel. For the molecules to create the gravity vector they must be aligned to the gravitational body. You can have one or the other, but you can't have both at the same time when there are multiple sources to align to.

I think what you’re describing here is what happens more or less strongly in the various states of the matter.

I’d like to remind to everyone, when studying the details don’t forget the whole picture.

When you talk about a particle like the proton, remember that it is formed by the stacked-spinning photon, yes, but also by all the channeling photons inside.
It is traveling at near-c linearly and it’s immersed in a photon wind that usually has a general direction.

So ultimately the proton is a definite perturbation of the photon wind it travels through. It is expected for it to be aligned by bombardment to a certain position.

There are some problems I’m still struggling with and probably you already solved them, so let’s just help each other. Some of these problems are the reason why a stacked spin is a stable cycle and not just a random wobble and whether (and why) incoming charge cones are formed at the nucleons poles.

Also the most basic problem, what is the photon, this something that always travels at c and it’s indestructible because it’s made of nothing else. But I expect this to never be solved.

Getting back on topic, I remember some time ago we talked about the pyramids, now it could be inferred that if they manage to disrupt the uprising charge direction, something on top or above them would feel less gravity. Maybe something more efficient and in a smaller scale can be designed and built to achieve the same result. Any ideas?

LongtimeAirman wrote:Nevyn wrote. Our moon is much farther away than our charge-pause.
Airman. As you can see from the attached image, that statement isn’t entirely true; on average, yes, as the caption says - the moon spends about six days a month in earth’s magnetotail. Those six days aren’t simply inside or outside the earth’s ion exclusion zone; the moon penetrates the magnetosphere “walls” twice each month. The moon can receive a great deal of protonic matter originating from both the earth or sun in each passage through the earth’s magnetosphere.

But it still spends more time in the sun's field than the earth's. So, at best, it will align to the sun and then, over those 6 days out of 28, it will start to adjust itself to the earth's field, but it will not complete that adjustment. These changes take time to build on such a large mass. Although I am thinking in changes to orbits, axial rotation or axial tilt, etc. So that isn't really fair. We are talking about nuclei alignment, so maybe we can take the time out of it and assume that, at least a fair few, nuclei will re-align to the earth. That gives us 6 days out of 28 that we can explain gravity of the earth on the moon. I guess we just aren't allowed to ask about it on the other days.

LongtimeAirman wrote:The celestial sphere – including the sun - spins about the moon roughly every 24 hours. That's sufficient for charge binding as is it is true between the spinning earth and sun. A big difference is the case of the earth and moon is the fact that Luna always presents the same face to the Earth. The earth sits, spinning in the sky once every 24 hours, but it never changes its spatial location with respect to the moon.

No, the celestial sphere, from the perspective of the moon, rotates around it every 28 days. Maybe you meant the earth?

LongtimeAirman wrote:The moon is in a molecular-like coherent alignment with the earth (aside from the earth’s rotation) which has increased in strength over time. Miles has pointed to the blasted face of the moon as due to charge from the earth. The flattening enables a broader charge binding linkage between the earth and the moon’s core, no doubt allowing a greater earth/moon orbital distance.

The moon is in so-called gravity-lock with the earth, not molecular alignment. I think the molecules can change alignment much quicker than the moon as a whole can. I would explain the gravity-lock through magnetism. Magnetism is the only force available to cause axial rotation of orbiting bodies. The moon feels magnetism from both the sun and earth, but the sun overpowers the earth's charge, at the orbital position of the moon. The earth's charge still provides a push, through its electrical component, and it also provides some spin through its magnetic component. But the sun also provides those things, with more strength. So the sun's field is wanting to spin the moon one way, but the earth's wants to spin it the other. This equates to a slow spin of the moon that happens to match its rotation around the earth. I'm sure there is some mathematical way to relate them, but I don't have it.

LongtimeAirman wrote:The sun’s field is far more extensive than the earth’s field, but at the earth/moon – sun radius, the sun’s emission field doesn’t get much stronger than the earth’s. Even if it did – to some degree, the sun is in constant orbital motion and so the sun can never establish the net channel/repulsion ratio with the moon, as the moon shares with the earth.

But it does. We know that the sun's field is stronger than the earth's at 11 earth radii from its surface. That distance is enough to negate the earth's magnetic field and the sun's to still exist, as a measured value. The moon is at 60 earth radii, so how can you suggest that the sun's field isn't much stronger at that distance?

I'm not sure what you mean about the sun being in constant orbital motion. Of course, it is, from a galactic perspective, but we aren't going that far. We aren't using the galactic core in this problem. If we did, then everything just moves down a step. The galactic core becomes a star, the sun becomes a planet, the earth becomes a moon, etc. Not in a literal sense, but they replace them in the above descriptions.

LongtimeAirman wrote:Nevyn wrote. Where are the outside photons coming from? The other planets, you say?
Airman. With respect to the moon, I believe the charge received from the sun is at a fairly constant (+/- the orbital radius) rate and magnitude.

Yes, the sun provides a fairly constant field, but what it doesn't provide is the right directional component, most of the time. The sun has a definite location with respect to the moon, at all times. That direction is imprinted on its charge field. Over the 28 days it takes the moon to rotate on its own axis, it will receive that charge over its surface, but gravity doesn't work over 28 days. It works at all times, in all positions. We can't save up the charge for later. We can't have the outside charge entering the nuclei and waiting for the inside charge to arrive and then spin-up and continue on its journey. We can't store the charge just behind the moon until we want it.

LongtimeAirman wrote:On average, the Earth’s field density equals the sun’s, but the moon experiences drastic changes in the earth’s charge field density when going through the earth’s tail.

You can't work in averages like that. You can average in some ways, but not this one. You can average the charge in the plane orthogonal to the line between the earth and moon, but you can't average it along that line, because to do so removes the gradient of that field. It just ignores the density differences that the distance creates. It flattens out the field, which ignores the properties of that field.

LongtimeAirman wrote:My general observation, the moon receives charge from both the sun and earth. It seems that some of the charge received from the sun – ok, or from the other planets too, will be diverted to the moon’s earth link. It seems to help explain the huge orbital radius. How does that affect gravity?

I don't think we need to bring other charge sources in to explain that distance. It only needs the electrical component of the earth's charge field to explain it. The distance is the balance of emission and gravity, whatever that mechanism may be. I don't think it matters how it is explained, for this problem, it is what it is. Gravity needs to work with that distance. The distance is irrefutable, the mechanism of gravity is not.

LongtimeAirman wrote:Nevyn wrote. Do we see a great increase in the earth's gravity applied to the moon when it is inline with the sun? Surely it has more photons on the right trajectory to create gravity in that position. It has the full force of the sun behind it. It even removes the alignment problem. I would expect a pretty big difference.
Airman. We do see a change in earth’s gravity during eclipses as shown by pendulum tests; I don’t agree that we should expect a pretty big difference. The coherency between celestial bodies never varies that greatly. I don’t believe you can add or subtract gravity as the planets pass in-line with each other as you seem to be suggesting. The moon should have been turned by the earth’s magnetotail wall during some grand maximum conjunction and lost long before now if it did.

Yeah, I was thinking about those differences as I wrote that, hence my last statement that I expect a large difference. Those pendulum experiments are showing minor variations in gravity. But if the moon suddenly has more charge coming from behind it, and the nuclei are already aligned, then surely that means that it is more likely to cause collisions between the sun's charge and the earth's charge at the moon. Everything is inline, the fields are not orthogonal to each other, so we have the perfect conditions for binding gravity to manifest. So I would expect some pretty large variations as compared to the other cardinal positions.

I'm not adding and subtracting gravity at all, I am showing a scenario that allows binding gravity to be expressed more, given my understanding of the mechanism. More inline charge, more inline nuclei, more collisions, more spin-ups, more gravity. But it is actually assumed that you can add and subtract gravity. The mainstream do it all the time. I recently posted about expansion stating that you can't do that with that theory. You can't add the expansion of different bodies to get an expansion of the collection of bodies. But the mainstream add up all the mass in a solar system, or galaxy, and just assume that the sum can now be used to calculate the gravity of the whole system. They also assume subtraction, but in a different way. They assume that the gravity of the earth and the moon are equal at some point and therefore, there is no gravity at that point (with respect to those bodies). And even though those mechanisms contradict each other, they do it anyway.

Ciaolo wrote:I think what you’re describing here is what happens more or less strongly in the various states of the matter.

Yes, it is combining a few different fields that we haven't really brought together before. We may be making some bad assumptions because of that. While I try to think these things through and find such assumptions, sometimes they hide very well. That's why I need you guys. I want you to show my misunderstandings. I want you to make me explain my position better, and possibly find them myself by doing so. I really appreciate the discussion. I don't really see it as an argument, because we all win, either way. Either I am right, and I have the correct understanding, or I am wrong and am shown to be, so now I have a better understanding. How can I lose? If I come out with a better understanding, then I win, even if I was wrong. This is not a zero-sum game.

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LongtimeAirman wrote:Nevyn wrote. Our moon is much farther away than our charge-pause.
Airman wrote. As you can see from the attached image, that statement isn’t entirely true; on average, yes, as the caption says - the moon spends about six days a month in earth’s magnetotail. Those six days aren’t simply inside or outside the earth’s ion exclusion zone; the moon penetrates the magnetosphere “walls” twice each month. The moon can receive a great deal of protonic matter originating from both the earth or sun in each passage through the earth’s magnetosphere.

Nevyn wrote. But it still spends more time in the sun's field than the earth's. So, at best, it will align to the sun and then, over those 6 days out of 28, it will start to adjust itself to the earth's field, but it will not complete that adjustment. These changes take time to build on such a large mass. Although I am thinking in changes to orbits, axial rotation or axial tilt, etc. So that isn't really fair. We are talking about nuclei alignment, so maybe we can take the time out of it and assume that, at least a fair few, nuclei will re-align to the earth. That gives us 6 days out of 28 that we can explain gravity of the earth on the moon. I guess we just aren't allowed to ask about it on the other days.  
Airman. Actually, in this case, the magnetosphere doesn’t matter. There’s no realigning. All celestial bodies align to all other nearby (angular wise) sources at all times. The moon doesn’t re-align receipt of charge from one source to the other. The moon is always receiving direct photons from both the sun or earth at all times.

Wherever the moon travels, the degree of charge it receives - magnitude wise - from the sun is roughly equal to the charge it receives from the earth; the system is in balance, the moon shares the same band of solar variability as the earth.

The earth’s magnetosphere is not a wall – if the moon should be so unfortunate enough to find itself inside – the magnetosphere cannot prevent the moon from receiving direct photon charge from the sun. Nor does the moon, when on the other side of the solar side of the magnetosphere “boundary” see any difference whatsoever in the receipt of coherent charge from the earth.
 
For all intents and purposes, the earth’s magnetosphere doesn’t affect the amount of photon charge the moon receives from the sun. The moon is always aligned with both the sun and earth at all times.

LongtimeAirman wrote:The celestial sphere – including the sun - spins about the moon roughly every 24 hours. That's sufficient for charge binding as is it is true between the spinning earth and sun. A big difference is the case of the earth and moon is the fact that Luna always presents the same face to the Earth. The earth sits, spinning in the sky once every 24 hours, but it never changes its spatial location with respect to the moon.

Nevyn wrote. No, the celestial sphere, from the perspective of the moon, rotates around it every 28 days. Maybe you meant the earth?
Airman. Thanks, you’re right. I meant to give that more thought. It still doesn’t alter the fact that the earth sits in one spot in the moon’s sky, while the sun is orbiting through that same sky. I imagine that singular constant direction allows the creation of charge binding to a stronger degree than the moon can establish with the angularly moving sun.  

LongtimeAirman wrote:The moon is in a molecular-like coherent alignment with the earth (aside from the earth’s rotation) which has increased in strength over time. Miles has pointed to the blasted face of the moon as due to charge from the earth. The flattening enables a broader charge binding linkage between the earth and the moon’s core, no doubt allowing a greater earth/moon orbital distance.

Nevyn wrote. The moon is in so-called gravity-lock with the earth, not molecular alignment. I think the molecules can change alignment much quicker than the moon as a whole can. I would explain the gravity-lock through magnetism. Magnetism is the only force available to cause axial rotation of orbiting bodies. The moon feels magnetism from both the sun and earth, but the sun overpowers the earth's charge, at the orbital position of the moon. The earth's charge still provides a push, through its electrical component, and it also provides some spin through its magnetic component. But the sun also provides those things, with more strength. So the sun's field is wanting to spin the moon one way, but the earth's wants to spin it the other. This equates to a slow spin of the moon that happens to match its rotation around the earth. I'm sure there is some mathematical way to relate them, but I don't have it.

Airman. A link to a permanent spot in the sky is what I was talking about. Gravity lock, eh? For the sake of this discussion it’s should be called charge binding. Given a molecular interpretation I see you have no problem comparing the relative strength/effects of simultaneous solar and earth charge. “The moon feels magnetism from both the sun and earth, but the sun overpowers.” Again, I disagree. With respect to the moon, neither the sun nor the earth overpower the other. They are both roughly equal, with charge from either body received at different lunar longitudes.
 
LongtimeAirman wrote:The sun’s field is far more extensive than the earth’s field, but at the earth/moon – sun radius, the sun’s emission field doesn’t get much stronger than the earth’s. Even if it did – to some degree, the sun is in constant orbital motion and so the sun can never establish the net channel/repulsion ratio with the moon, as the moon shares with the earth.

Nevyn wrote. But it does. We know that the sun's field is stronger than the earth's at 11 earth radii from its surface. That distance is enough to negate the earth's magnetic field and the sun's to still exist, as a measured value. The moon is at 60 earth radii, so how can you suggest that the sun's field isn't much stronger at that distance?
Airman. I need to work this argument up with numbers. Of course, everything is in balance, the long earth/moon orbital distance is central to the discussion.

Nevyn wrote. I'm not sure what you mean about the sun being in constant orbital motion. Of course, it is, from a galactic perspective, but we aren't going that far. We aren't using the galactic core in this problem. If we did, then everything just moves down a step. The galactic core becomes a star, the sun becomes a planet, the earth becomes a moon, etc. Not in a literal sense, but they replace them in the above descriptions.
Airman. The sun orbits through our sky, and the sun orbits through the moon’s sky – every 28 days, thank you. There is a big difference, the earth sits at a stationary spot in the moon’s sky. The stationary position allows the creation of more binding material than is possible for the moon to establish with the orbiting sun.

LongtimeAirman wrote:Nevyn wrote. Where are the outside photons coming from? The other planets, you say?
Airman wrote. With respect to the moon, I believe the charge received from the sun is at a fairly constant (+/- the orbital radius) rate and magnitude.

Nevyn wrote. Yes, the sun provides a fairly constant field, but what it doesn't provide is the right directional component, most of the time. The sun has a definite location with respect to the moon, at all times. That direction is imprinted on its charge field. Over the 28 days it takes the moon to rotate on its own axis, it will receive that charge over its surface, but gravity doesn't work over 28 days. It works at all times, in all positions. We can't save up the charge for later. We can't have the outside charge entering the nuclei and waiting for the inside charge to arrive and then spin-up and continue on its journey. We can't store the charge just behind the moon until we want it.
Airman. Agreed. Gravity is a constant process. As Vexman indicated we are talking about very quick responses, but it is a process. The charge bound nuclei must constantly reorient or trade off with other adjacent matter in a slightly better vantage given an asmuthally moving sun. It's easier to align atomic matter with the earth, since the earth is always at the the same approximate lunar longitude.   

LongtimeAirman wrote:On average, the Earth’s field density equals the sun’s, but the moon experiences drastic changes in the earth’s charge field density when going through the earth’s tail.

Nevyn wrote. You can't work in averages like that. You can average in some ways, but not this one. You can average the charge in the plane orthogonal to the line between the earth and moon, but you can't average it along that line, because to do so removes the gradient of that field. It just ignores the density differences that the distance creates. It flattens out the field, which ignores the properties of that field.

Airman. Pretty much the only effect I would expect the moon would experience as a result of the moon’s transit across the earth’s magnetosphere is a higher than average density of proton matter when passing through the magnetotail walls; although wall is a complete misnomer - more like a mist or a veil of smoke.

LongtimeAirman wrote:My general observation, the moon receives charge from both the sun and earth. It seems that some of the charge received from the sun – ok, or from the other planets too, will be diverted to the moon’s earth link. It seems to help explain the huge orbital radius. How does that affect gravity?

Nevyn wrote. I don't think we need to bring other charge sources in to explain that distance. It only needs the electrical component of the earth's charge field to explain it. The distance is the balance of emission and gravity, whatever that mechanism may be. I don't think it matters how it is explained, for this problem, it is what it is. Gravity needs to work with that distance. The distance is irrefutable, the mechanism of gravity is not.
Airman. That's Ok with me, it wasn’t my intention to bring in other sources, but not to leave one of your comments out.
 
LongtimeAirman wrote:Nevyn wrote. Do we see a great increase in the earth's gravity applied to the moon when it is inline with the sun? Surely it has more photons on the right trajectory to create gravity in that position. It has the full force of the sun behind it. It even removes the alignment problem. I would expect a pretty big difference.
Airman wrote. We do see a change in earth’s gravity during eclipses as shown by pendulum tests; I don’t agree that we should expect a pretty big difference. The coherency between celestial bodies never varies that greatly. I don’t believe you can add or subtract gravity as the planets pass in-line with each other as you seem to be suggesting. The moon should have been turned by the earth’s magnetotail wall during some grand maximum conjunction and lost long before now if it did.

Nevyn wrote. Yeah, I was thinking about those differences as I wrote that, hence my last statement that I expect a large difference. Those pendulum experiments are showing minor variations in gravity. But if the moon suddenly has more charge coming from behind it, and the nuclei are already aligned, then surely that means that it is more likely to cause collisions between the sun's charge and the earth's charge at the moon. Everything is inline, the fields are not orthogonal to each other, so we have the perfect conditions for binding gravity to manifest. So I would expect some pretty large variations as compared to the other cardinal positions.

I'm not adding and subtracting gravity at all, I am showing a scenario that allows binding gravity to be expressed more, given my understanding of the mechanism. More inline charge, more inline nuclei, more collisions, more spin-ups, more gravity. But it is actually assumed that you can add and subtract gravity. The mainstream do it all the time. I recently posted about expansion stating that you can't do that with that theory. You can't add the expansion of different bodies to get an expansion of the collection of bodies. But the mainstream add up all the mass in a solar system, or galaxy, and just assume that the sum can now be used to calculate the gravity of the whole system. They also assume subtraction, but in a different way. They assume that the gravity of the earth and the moon are equal at some point and therefore, there is no gravity at that point (with respect to those bodies). And even though those mechanisms contradict each other, they do it anyway.
 
Airman. Uncle!

Thanks for the attention and extensive reply. Aside from adding and subtracting gravity I think I appreciate all the difficulties you are describing. On the other hand, all the planets are linked to the sun. The earth links with both the sun and the moon. I don’t understand why you seem to think a celestial body could only link to one other body, or that the links should strongly interact.

This discussion has been a dandy. I feel like everyone’s learned something, even considering the worst.

Ciaolo, don't lose hope, I'll try to answer your comments tomorrow.

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I'm happy with the bodies being linked to different sources via charge. I'm happy with that charge effecting that body. I just can't accept that the nuclei of that body can be aligned to more than 1 at any given time. I am only saying that it is binding gravity that requires that alignment, and I can't see how it can maintain that alignment with multiple sources of charge to align the nuclei. I am not saying that alignments can't happen, or that the moon, say, only feels the charge of the earth or only the sun. It feels both at all times, and it balances all of the forces that it feels which we see as its orbit. Happy with that. It is how binding gravity operates that causes problems, at least as far as I see it at the moment.

I'm not even saying that binding gravity is a total loss, although I would forgive anyone for thinking so. There may be ways around what I have found, or I am just straight up wrong about some or all of it. I'm always open to that possibility, but it might take some effort to convince me. I argue hard and I argue deep, at least I do when I believe I have a decent understanding of the issue. I thank everyone for giving their opinions on these hard topics. I hope I haven't come across as harsh. I try to keep my posts devoid of such emotions, but may not always be successful. I certainly don't want to stop anyone from saying what they think about it. Keep up the good work, and keep on thinking!

Nevyn wrote:The moon feels magnetism from both the sun and earth, but the sun overpowers the earth's charge, at the orbital position of the moon. The earth's charge still provides a push, through its electrical component, and it also provides some spin through its magnetic component. But the sun also provides those things, with more strength.

Nevyn, this suggestion is exactly where you and I had different opinions. You said that at the charge pause zone, solar wind strips off the magnetic component of Earth's charge field (at 11th ER). And you further said that it's possible we have some Earth's charge field remaining after the charge pause, but only with its linear component left. Actually, we both agree that the charge field doesn't completely stop at 11th ER. But still, the essence of your rebuttal while replying to me was paraphrasing Miles, that the charge pause limit represents a point where those two opposing fields are equal in strengths. You weren't even allowing the remaining Earth's field to bear any significant density and/or presence beyond this limit, which opinion I opposed equally.

Now, you shouldn't allow Earth's charge or its linear component reaching Luna if you want to remain consistent to your rule set as paraphrased just above. There is no significant Earth's magnetic field beyond 11th Earth radii if you stick to your own explanation.

But we're leaving out Luna's E/M field and underlying charge field, completely forgetting its presence here. Her field certainly deflects some Sun's charge and she helps Earth deflecting/blocking some charge when in the right celestial position (when passing between the Sun and the Earth). Luna and it's field certainly have its role in the balance of orbital forces. Maybe the reason she doesn't rotate on its axis is connected to the gravitational lock to Earth and the fact its submerged in much stronger field set by the Sun - she is constantly switching her alignment and that makes her look static, locked in in the imperfect angle to either of the bodies fighting for the field's dominance. So only one Lunar position allows the correct balance.

I was sloppy with my words and didn't explain enough. I hinted at certain things that I assumed were obvious. I was referring to field strength, not existence. Beyond the charge-pause, the sun is stronger, but the earth still reaches to the moon to provide the repulsion required to explain orbits. I stated that the charge-pause is not the point where charge stops. It is just the point of equal strength. Both fields reach beyond that point, or maybe I should say charge. The sun's charge reaches all the way to the surface of the earth.

If a photon emitted by the earth makes it to the moon, then it did so because it didn't collide with anything in between. Therefore, it retains any spin that it had when leaving the earth. That spin creates magnetism. It isn't magnetism, but it can create it. So the magnetism of the earth does reach the moon, in a way, but it is not measurable and as such, it isn't the magnetic field. The magnetic field is measured by probing various locations and recording the values. It is like dropping a ping-pong ball in a stream and recording which way it goes. Then moving to another location and doing it again.

What we call magnetism is the action of the spin of the photon when it collides with something. It is the resultant motion of that which it collides with. So in this way, the charge-pause is where the earth's magnetic field stops, because the action outside of that radius is a result of the sun's magnetism. The sun's field overpowers it. That doesn't mean the earth's field is not present or affecting the motion as well. It just isn't enough to care about. We don't really see it in quick measurements of magnetism.

When talking about axial rotation, though, we are not using quick measurements. We are using very large time spans, where the fields can apply pressure. The moon is being struck by the earth's charge which only needs to push it away to explain orbits. For axial rotation, we just realise that those charge photons must still be spinning, and enough of them over time will create a measurable effect. Then, as Airman showed us, we have periods when the earth has more sway than the sun. It is all a dance between the charge fields of the three bodies. These kinds of forces take time to build, though.

Our discussion about the alignment of nuclei does not have great time spans to work with. In fact, it has very little time, if any. If it did take time, then we would notice variations in gravity, but it seems to be fairly constant. We would also notice a reaction time to gravity, which has never been shown. It seems gravity either works at or above c, or it is infinite in speed (or instantaneous).

These concepts can be tricky to juggle. Things can be used in different ways when looking at different problems. What look like contradictions, can just be a different context, with different assumptions and expectations. See Miles' papers on axial tilt and axial rotations, which are probably in the third wave papers, I think.

I'm not nit-picking, here's what you've said with my addition in [brackets] or bold text. Using Airman formatting due to many quotes:

Nevyn: Let's imagine that they [emitted photons] reach another galaxy. Do we now say that the earth's charge field reaches that far? There has to be some practical limit as to what we call the earth's charge field and what has become just random charge photons. I suggest that that limit is where one component or the other is lost. The charge-pause is where the magnetic component is lost. The individual photons have not necessarily lost their spins, but they are now mixed with other charge that has different spins, so the magnetic field drops to 0, or whichever is the stronger field. The electrical component can continue on and still be effective, but can we still call it the earth's charge field? Maybe we should call that the earth's electrical field, since the magnetic component has been overpowered.

You  [Vexman] are suggesting that the moon sets the limit for the earth's charge field, but it doesn't. That is the point that the earth-moon system matches in forces.

and following with:

Nevyn: I would say that we can still call the earth's emission a field for a little way past the charge-pause, because it still has some appreciable density and structure. It isn't dominant, but it still has an appreciable effect. Can we take that all the way out to the moon though? I don't think so. It is over 5 times the distance of the charge-pause, where we know that the sun is equal. Maybe we could allow another 11 earth radii, just to be generous, to that and say that the earth's charge field extends to 22 earth radii from the surface.


So, in your last post and starting with sparse Earth's emitted photon reaching Moon you conclude about the magnetic field:

Nevyn: The sun's field overpowers it [earth's magnetic field]. That doesn't mean the earth's field is not present or affecting the motion as well.


Which one is correct since we can't have both? Do we have Earth's charge field reaching Moon or not? Sparse photons can't be thought of as a charge field, you told me this yourself and it plays a crucial role here. So if emitted charge is actually a charge field required to explain orbits, which we know it is, than this confirms the thesis I support - the charge-pause is only about the magnetic field boundary, while the charge field (stripped of magnetic force - spin) reaches to the Moon's orbit. So this leaves the question opened about Miles' own claim where he says , with my bold text:

Miles: "The magnetic field would not be expected to extend beyond the charge field, because the charge field creates the magnetic field. And the magnetic field would not be expected to stop short of the charge-pause, because there is no reason for it to stop short. Ions are driven up by the charge field, and will persist as long as the charge field persists."

No, I think the correct sentence would be: And the magnetic field would not be expected to stop short of the charge-field, because there is no reason to stop short.
But it does, doesn't it? Magnetic component stops at the charge-pause because of the solar wind, while charge field keeps its direction and linear component and reaches Moon, giving it enough push to result in a steady orbit. So we have 2 different issue here at hand: the phenomenon of field's magnetic component loss at charge-pause and the phenomenon of "elusive" field which extends beyond charge-pause. So if charge-pause is about the point of Sun's and Earth's equal fields' strengths, what is the Moon's orbit about? If the former explains the loss of field's magnetic component, what can explain the latter?

Yes, I admit that I used certain terms in certain ways, then remembered other things as the discussion progressed, or even sometimes as I was writing. I always allowed the earth's charge to reach the moon in order to exclude it for orbital purposes. Maybe I didn't state it explicitly enough when I should have. I later used those photons, containing a residual amount of magnetism, to explain the axial rotation (which comes straight from Miles and the theory we are working with).

The discussion was about aligning nuclei, not rotating solar bodies. A nuclei can adjust to the current fields in a very quick time frame. A large body can not, because it has a hell of a lot more mass to move. This means it takes time for those forces to build up to a measurable amount. It is a completely different problem, although related at the causal level.

Then there is the definition of magnetism, which I tried to explain in my last post. The magnetic field in a given location is the result of all fields at that location. The strongest will provide the most influence. It may be stronger because it is more dense, because it has more mass (more spins on the charge photons), or a combination of them. So the earth's magnetic field is a different thing to the magnetic component of its charge field.

I did explicitly state that it lost the magnetic component, and that was bad wording on my part. I should have said the earth loses its magnetic field, so that it didn't sound like the spins were being stripped from the charge photons. I never meant that and stated so, it's right there in your quote of my statement. I always expected some charge to make it to the moon and I always expected it to retain its spin. I didn't expect it to overpower the sun's charge, as a measurement of magnetism, but the force is still there if the earth's charge photons are there and no matter how small it is.

When I quoted Miles about the extent of the charge field, I actually went looking for that to correct myself. I remembered that Miles had talked about it and that it would effect what I was writing at the time, so I checked and found that he was saying what I was saying. But at other times, say when discussing the orbit of the moon, Miles talks of the charge field as reaching the moon. This is because the charge field is being used in a different way. The focus is on different aspects of that field.

So you can sometimes say that the earth's charge field reaches the moon, and in some ways it always does. But when you have multiple fields in the problem, you have to be clear about the precedence of those fields at each individual location. You also have to be clear about what it is effecting and how it is doing that. The effects on a large mass will be different from the effects on a very small mass. A large mass sort of averages out forces over time, where-as a small mass reacts quickly to the current forces.

So I stand by both statements, while admitting that I didn't explain it very well, although in my own defense, I was focusing on the problem at hand and not muddying the waters with every other application of the entities we were discussing. I expected you to understand that any photon that makes it to the moon did so because it avoided collision, so it must retain its own individual spin. The field has lost strength, in comparison to the other fields at a given location, but it has not lost existence. Each individual photon has not lost its own strength, but the collective has. That is the difference between talking about a field and talking about photons. A field is a collective. It is not individual photons, it is the result of those photons acting upon other entities.

So, strangely enough, we can say that the magnetism is lost, or more correctly overpowered, while still retaining some of it to reach the moon. The field has been overpowered by the sun's field, because that field has more strength so it effects entities more, but the photons that make it to the moon have not been overpowered because they have not collided with anything to change their spins. But that residual magnetism is small. It still exists, but it needs time and variance to build up to a measurable resultant motion on the moon. By variance, I mean there is some amount of time where the earth's charge has a stronger influence than normal.

I understand the confusion, and I have no problem with you pulling me up on it. I often notice these things as I write and chuck in a sentence or two to show some minor quality, like this:

Nevyn wrote:The charge-pause is where the magnetic component is lost. The individual photons have not necessarily lost their spins, but they are now mixed with other charge that has different spins, so the magnetic field drops to 0, or whichever is the stronger field.

As I wrote the first sentence, I realised how it would sound and tried to show that the individual spins are not lost in the next sentence. I was also trying to differentiate between the concept of magnetism and the spins on the photons. But the main focus was about whether we could say that the nuclei of the moon could be aligned by the earth's charge field, which means we were discussing the relative strength of multiple fields, not individual photons in an isolated scenario. I was trying to show that the sun's field is stronger, so the nuclei would align to that and not the earth's field.

In physics, or any problem solving, you have to be very careful of the context of the problem at hand. Concepts can take on different meanings in a different context or different way of looking at a problem. I am a Software Engineer, so I deal with problem solving in a general sense. We are often solving many different types of problems in many different ways. We become accustomed to seeing solutions without any context, and then applying them to a specific context. This allows us to be very abstract and to see things in many different ways. We, or at least some of us, understand perspective a bit better than most, I would say. I often take that for granted.

So when trying to understand a given problem, you have to sort through the ways that certain concepts will manifest. Sometimes you will look at it from a broad perspective and sometimes you will look at it from a very narrow perspective. The exact same actions can look quite different from those two perspectives.

For example, if we are looking at individual photons, a very narrow perspective, then we might say that each photon has its own magnetism. Strictly speaking that is incorrect. They have spin, which can create magnetism. When looking at fields, a broader perspective, then magnetism is the result of many photons.

In the case of an individual photon, it is either spinning or it is not. We can't ignore that spin in a collision. But when talking about fields, then individual photons can still apply their spin, but other photons can overpower it. We are talking about many photons, not individuals. We also have more time when discussing a field, so we have pressure, rather than force alone. A field is many collisions, but a photon is only a single collision.

I'm not sure if I am explaining that very well. I'm just trying to say that you can't always create a single definition of a given concept and apply it to all scenarios. You will run around in circles trying to catch your own tail. You have to figure out how a given concept applies in the situation you are dealing with at the time, in the way that you are dealing with it. There are many ways to skin a cat, as they say (sorry Miles).

I've been through this myself. I thought I had found contradictions in Miles papers and got all confused about everything. As I tried to work my way through it, I soon realised that the contexts were different and I had to adjust my own understanding of certain concepts in order to give them enough freedom to be expressed for different problems and to be expressed in different ways.

It is easy to read through Miles papers and come to the conclusion that magnetism is the spin on charge photons. Both he and I are guilty of calling it that at one time or another. But deep down, we both understand that it isn't really true, and have stated so, but not necessarily every time we use it. In some ways it is laziness, but it is also about being focused and sticking to the current problem without needing to re-write the entire theory out every time we use a specific part of it. My posts are long enough without doing that.

Because of my profession, I am used to switching perspectives (or maybe I am in that profession because I can do that?). I often need to have multiple ways of seeing something in my head and evaluate them to determine which one I will use. Physics is the same, or at least theoretical physics is. You often need to juggle multiple models and find the differences. When you have done this enough, you start to see things in different ways. You start to see things more abstractly and you start to notice abstractions in other places. It is a very handy skill.

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Hey Ciaolo, I hope you don't mind, I'm going to be as nit-picking precise as possible.
 
Ciaolo wrote. In my opinion the most important aspect of this new theory is that we go beyond simple pool ball mechanics.
Airman. Agreed. Miles addresses this issue directly in The Cause of Gravity the next major chapter  http://milesmathis.com/grav3.pdf.
Miles wrote. So you can already see that this isn't strictly a poolball problem, although I love those. I have been selling poolball mechanics for almost twenty years, to counter the rise of mysticism in mainstream physics, so you can see why I was fooled by this one as well. We will keep it mechanical, but it isn't a naïve poolball mechanics. It is a charge mechanics.

Airman. In this thread above, BorisT asked for a simple 'pin-ball mechanics' description, I answered by saying "I believe Miles pointed out that ‘pool ball mechanics’ might be a bit oversimplified. I would suggest changing it to ‘spin-ball mechanics'". At the moment, it didn’t help. Upon another re-read I noticed that later in the paper, Miles makes a better suggestion.

Miles wrote. we don't have a naïve vector mechanics of straight collisions here, we have a complex spin mechanics.

Airman. Bold, extra-size, all-caps (too rude). Spin mechanics. I would urge anyone with their own pool/pin/spin ball terms, (I’ve got more than one) please refer to spin interactions as Spin mechanics.

Ciaolo wrote. Earth emission up is not only a bombardment that would push, it is actually a components of 2 types of photons. A component can be channeled and the other cannot. Only the latter causes a push.
Airman. Agreed, although types is confusing. Two types of photons: 1. Can be channeled and 2. Cannot be channeled. That’s not true. Given the right direction, any photon can be channeled; a photon is either channeled or not. Or most photons avoid both those possible outcomes as well as the atmosphere and make it to space. What we are really talking about the interaction of atomic matter and the earth’s emission field. So, to avoid that confusion.

Upward earth emission photons ‘colliding’ with matter either: channel into; or repel that matter.  

Ciaolo wrote. The same applies to the charge field the Earth is in, part of it can be channeled and part cannot.
Airman. What part of charge cannot be channeled? Again, charge that collides with the earth that is not channeled causes a repulsion.

Ciaolo wrote. Of course if a body is not spherical all of that charge cannot be channeled so it becomes a push.
Airman. Ok, but you’re losing me. Any photons colliding with the matter that are not channeled cause a push.

Ciaolo wrote. In my opinion 9.81 is both the earth emission and the Earth local ambient field. The resulting 9.80 is the ratio between the recycled flux up and the bombarding flux down. But this is only my thought, I still haven’t polished this enough to explain it thoroughly, let alone proving it to be correct.
Airman. I’m not happy with “bombarding flux down“. With respect to earth’s upward emission, and its interaction with atomic matter we find photons either channeling or repelling matter. Where did the bombarding flux down come from? I do not believe there is a bombarding downward flux. My understanding is that charge channeling is charge binding.

Ciaolo wrote. I’d like to remind to everyone, when studying the details don’t forget the whole picture.
Airman. Absolutely.

Ciaolo wrote. When you talk about a particle like the proton, remember that it is formed by the stacked-spinning photon, yes, but also by all the channeling photons inside.
Airman. Agreed. The only problem is, we don’t have all the details figured out yet.  

Ciaolo wrote. It is traveling at near-c linearly and it’s immersed in a photon wind that usually has a general direction.
Airman. Agreed – although, don’t you think a proton traveling near-c is a bit fast, Solar wind? It’s got to stop somewhere. Usually, compared to photons, most proton matter can be considered stationary.  

Ciaolo wrote. So ultimately the proton is a definite perturbation of the photon wind it travels through. It is expected for it to be aligned by bombardment to a certain position.
Airman. I would change “the photon wind it travels through” to “the photon wind both aligns and pushs the proton.”

Ciaolo wrote. There are some problems I’m still struggling with and probably you already solved them, so let’s just help each other. Some of these problems are the reason why a stacked spin is a stable cycle and not just a random wobble and whether (and why) incoming charge cones are formed at the nucleons poles.
Airman. I bolded two questions you indicate as problems. In all honesty, you know we haven’t solved those questions. In my opinion, the charge field has been shown to be true, so while I cannot give satisfactory answers to those questions, I think we’ll figure it out. That is, if Miles doesn’t eventually answer every question himself.

If I weren’t discussing this topic with you, I would probably be involved with the possible charged particle field particle engine. I have given particles a great deal of thought. Therefore, I’d defer the first question to Nevyn.

I‘ll try answering the second question - whether (and why) incoming charge cones are formed at the nucleons poles. I hadn’t given incoming charge cones - vortices - any thought until Miles began referring to them in his papers. I don’t recall him ever describing how they form, but I kept thinking about it. I’ve made several vortices comments in this string and I can now say I am completely comfortable with the idea. I don’t know if I’m right or whether I’ve convinced anyone, but I’ll try laying it out for you to review.

The proton’s pole is aligned to the charge source below, channeling charge and setting the proton engine in motion. The proton is also recycling charge. A low pressure is generated in the nucleus creating two opposing incoming polar charge currents - upward and downward along the proton’s spin axis. Note, we don’t have vortices yet. Nor do we have the detailed physical model of the proton engine yet, this is all charge field theory. Please let me know where you disagree.

I would say the first requirement necessary for vortices to occur is that charge must be present in different sizes. Well that’s pretty much a charge field given. Miles has always described different energy photons with different sized radii. The various photon species extend across the entire electromagnetic spectrum – a huge range of size changes. Charge doesn’t need to be spherical, it can be toroidal, or a stacked spin bphotons, all the vortex model I’m describing needs is different sized charge. The second requirement would be to allow that large charge particles can have a low forward velocity.

The smallest charge, photons and antiphotons usually pass near each other along the spin axis without collisions. At the other size extreme, electrons cannot pass through the proton without collisions. These collisions leave the electron with little to no forward linear velocity, the electron loses energy to the proton, the electron has become smaller, it will take some time for the electron to be recycled. So the speed at which large charge can recycle through the nucleus is inversely related to the charge size, bigger particles take longer to recycle. Meanwhile, due to the proton’s e/m field, a migration of large, low velocity charge begins to assemble at the proton’s poles. The large charge particles are themselves recycling charge.

This is how I’ve been thinking about it. Please feel free to comment, question or correct me.  

Ciaolo wrote. Also the most basic problem, what is the photon, this something that always travels at c and it’s indestructible because it’s made of nothing else. But I expect this to never be solved.
Airman. I must agree.

Ciaolo wrote. Getting back on topic, I remember some time ago we talked about the pyramids, now it could be inferred that if they manage to disrupt the uprising charge direction, something on top or above them would feel less gravity. Maybe something more efficient and in a smaller scale can be designed and built to achieve the same result. Any ideas?

Airman. Miles’ idea of charge channeling/turning charge perfectly explains the way the Jean-Louis Naudin’s “lifter” works. It also explains helicopters and drones. That’s a topic or two from the Flying Saucers? https://milesmathis.forumotion.com/t453-flying-saucers thread. You can post there if you’ve got any flying ideas ,  it seems a little off-topic for this thread.
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I would also like to say that when I tried to limit the extent of the earth's charge field, I was doing exactly what I have cautioned about above. I was trying to create a single definition for all purposes. You have shown me what I already knew, but forgot, that that is a lost cause. We often need to define the charge field for the given problem. We can have a general idea of what it is and how it operates, but some things, like limits, can be different in different problems or even just different ways of solving the same problems. We need to be flexible. Not so that we can fudge any solution we want to, but to allow the concepts to be expressed in appropriate ways for various scenarios.

I do apologize for any confusion that I have created, but I believe my usage of these concepts is consistent at the mechanical level, even if it doesn't always look like it from various higher levels. We are quibbling about the definition of the charge field and where it does or doesn't stop. But it was never about stopping, it was about what it can do at various points between the earth and moon and how it relates to other fields. I was looking for a practical limit, which means that in practice it may not have an effect at some location, even though it is still present.

Ciaolo wrote:There are some problems I’m still struggling with and probably you already solved them, so let’s just help each other. Some of these problems are the reason why a stacked spin is a stable cycle and not just a random wobble and whether (and why) incoming charge cones are formed at the nucleons poles.

This is not the thread to be discussing this, but I'll throw in a quick answer and if anyone wants more detail then they should start a new thread for it.

The building of stacked spins begs two immediate questions: How can a spin be stacked; and How can it maintain that spin. You are asking about the second.

The first is inappropriately answered with the words: Spin Mechanics. Well, I guess both are really, but I'm just trying to say that the details are long and laborious. This is not the place to discuss how it stacks a new spin, other than to say that it requires the right sort of collision and it requires an existing spin (which can just be the axial spin). When the vectors line up in the right ways, then a new spin is stacked.

That isn't a very satisfying answer, but I'm trying to get to the actual question.

Maintaining spins is simple in one sense and complicated in another. In the simplest of cases, we just invoke Newton's laws of motion and state that a velocity will continue until acted upon by another force. So if we ignore everything except for our little spinning BPhoton, then it will spin forever, for there is nothing to stop it doing so. Please don't ask how it got those spins without any other photons around .

On the complicated side, we realise that the particle is not existing in isolation, and therefore it will collide with other particles, both big and small. That may not happen very often, or it might happen in rapid succession. In the case of a charged particle, it does happen often and regularly. So let's focus on that.

The first thing to realise is that a charged particle is a large entity in relation to the charge field. The top spin level of an electron is many spin levels above the top spin level of an infra-red photon. So in a collision between them, the electron effectively has more mass and the charge photon doesn't, so it can't completely remove that top spin. It can probably only reduce it a small bit. Of course, the converse it true too. If the electron's top spin level has been slowed down a bit, then another collision can increase it back up.

That is how the electron or proton can maintain its own spins while recycling charge. Charge comes in from all angles, mostly the poles but it can also come in at the equator. Even if we ignore the equator, charge coming in at the poles is bi-directional. So if charge entering the north can slow it down, then charge entering in the south can speed it back up. But it is actually more complicated than that, because charge from the north can also speed it back up when it is on the other side of the top spin cycle. That is, half the time it is moving up and half the time it is moving down during 1 spin cycle, so charge from either side can slow it down and speed it up.

With respect to charge vortices, I don't believe that they are vortices, but they might look like one. This is just the result of the way a charged particle pushes charge around. It prefers the equator to emit charge and prefers not to emit it at the poles. It still can and does emit at the poles, but the majority of charge is pushed to the equator. That is because the equator is in the same plane that the charged particle's top spin level is moving in. If the top spin level is a Z spin, then the motion is in the XY plane. Therefore, the equator will be in the XY plane too.

This creates a repulsion at the equator that limits incoming charge from this area. Therefore, the poles still receive the same charge they did without the particle being charged and it looks like it is pulling charge into itself. That is an abstraction, though, for it can not pull in charge. It just accepts what is already going that way. But the equator does not just accept anything moving its way. It is a bully, and tries to get rid of it before it can actually strike the charged particle itself.

So much for a quick answer.

I was wrong, I have my forces in the wrong planes. But the solution actually works better.

I was using the poles to supply charge to change the speed of the top spin level of a charged particle. But that is incorrect. Polar charge does have a chance to strike the BPhoton because that BPhoton does move along the Z dimension when it is at the center of the particle. However, that motion is caused by inner spins, not the top level.

So we actually need equatorial collisions to alter the top spin level of a charged particle because that is the plane of motion for that spin level. Therefore, the equatorial charge protects the top spin level. This may explain why it takes a similar sized particle to effect that top spin level. It needs some mass to get through the emission field. But even small charge photons can get through occasionally.

I should probably follow on from that and give a clearer picture of why the equatorial charge can be created without effecting the top spin level of the charged particle.

The reason is the direction of the forces. The equatorial emission is mostly created from the polar charge intake. The polar charge is moving along the Z axis, for a top level Z spin, but the Z spin is moving in the XY plane. So the polar charge can't affect the top spin level. This creates the equatorial emission in the XY plane by simpler, direction changing, collisions. However, charge that comes in from the equatorial region, so it is moving in the XY plane, can change the top spin level, but it must first get through the equatorial emission field.

The reason I wanted to clarify the issue of Earth's charge field presence first was because I had a sketch on my desk for a week or so, but couldn't present the idea since there was no Earth's field to work with at a distance. Now that we agree Earth's field is present and acting strong enough at the Moon's orbit/surface, I'd like to bring that idea forward.

Here's a picture I found to replace my sketch as it is much better done, although not perfect:



Now if you can imagine all those vectors having another vector drawn right next to them, but facing in the opposite direction (down) - that would be pretty accurate representation of what we experience on the surface of a gravitating body, i.e. the Earth.

The issue is orthogonality of binding force / gradient. It's always experienced as orthogonal to the Earth's surface and it is the same on both sides of the Earth, regardless of day-time or night-time (facing the Sun or not, in essence).

If true, orthogonality implies that the source (or more accurately, the angle) of incoming charge reaching the Earth is irrelevant. That also corresponds to our experience with binding force on Earth. Following the idea, if incoming charge is somehow aligned to collide orthogonal with Earth's emitted charge head-on, the source of charge becomes irrelevant in the sense, that its incoming (orthogonal) angle is now the same for any such source.

If true, orthogonality also requires a mechanical process of such alignment. I have a couple of ideas that sound reasonable in my thoughts, but would like to hear your comments about them.

The first option would be about the existing channels of Earth's emitted charge - it is assumed that the vector of repelling charge force is already orthogonal to the surface. Which means there are existing "virtual highways" or chanells of photons shooting out. Likewise, there is some space between those channels, conical in shape, where incoming charge tends to enter as the path of least resistence. A close analogy of this would be roughly like this:



The incoming charge enters this conical-shaped empty space and gets bounced of by hitting the upward photon, aligning itself more and more with each bounce. It is also accelerated while colliding with the same-spin emitted photons coming upward. With diminishing distance toward the emitting charge surface, the field is denser and denser, until incoming charge is almost perfectly orthogonal - a consequence of many collisions on its way down.

The second option would be Earth's E/M field (or magnetism?), which would give surrounding charge the correct spin at impact to turn them towards the Earth. I lack imagination to describe it properly, but maybe I can't imagine it because it isn't feasible. I remember Jared had criticized such idea, but I'd still like to explore this possibility - if E/M field has the capacity to deflect solar wind, maybe it can align the direction of incoming charge with similar mechanism. While E/M loses its strength (individual spins) to deflect, maybe it could get strengthened (stacking individual spins) to achieve a mirrored result - re-directing the charge towards the surface.

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Hi Vexman, thanks for joining in, great discussion twixt you and Nevyn.

Please consider the above image in light of yours. I was trying to maintain the proper scale between the Earth and Moon, but obviously the orbital distance isn't to scale with the earth and moon. Sorry about the apparent cutoff, as usual, larger images need to be copy pasted to where one could see the whole thing.

The large ends of both cones include all the direct emissions that reach the other body. I would like to point out that the small ends of the cones are providing small deflections to allow increased or the concentrated return charge that Miles has referred to with respect to Neptune's charge returning to the sun. The diagram isn't so much orthogonal, although it is; the diagram actually shows radial lines, (they just happen to be orthogonal too).  The concentrated returning charge is a big factor in the earth/moon system, and I believe this diagram conveys that.
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Here's another image which I believes shows the 'true scale'.

These numbers assume direct radial emissions. Most noteworthy to me is that rather small angular cones at the earth and moon encompass all the earth and moon's direct emissions.

This may be part of why I cannot believe celestial bodies completely align with each other. You can see why the sun can align with all the planets simultaneously, each channel is rather narrow. There's no way to expand that angle, other than say blasting away part of the surface as has occurred on the face of the moon. In my opinion only a small part of the proton matter in each celestial body is charge bound to each other.
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The problem I have with using the earth's charge field to somehow corral the incoming charge into the right trajectories, whether that be through simple collisions or through magnetism, is the same as I have already been arguing. The field has a limit to its strength in relation to other fields. If this were the mechanism to align the charge, then gravity would suddenly reduce in value at the charge-pause, just like the magnetic field does and for the same reason.

You can get some success using these ideas in a simple 2-body gravity problem. The problems arise when you have orbits within orbits. Or I really should say gravitational fields within gravitational fields. I say orbits because we actually need the orbiting body to see the gravity of the middle body (the earth in this case). We need the orbiting bodies because that is where gravity is being expressed. That is what gravity is working on, and in the case of binding gravity, working through. That is, binding gravity actually uses the nuclei of the orbiter as part of the gravity mechanism.

We know that the moon is orbiting the earth, yet the earth's charge field is not the dominant field at that distance. So any channels are going to point back to the sun. If we treat the channels as roads, then at the moon, the sun's channels are a highway, but the earth's are a dirt track in the bush (forest to non-Australians) at best. Not only are we dealing with the sun's charge creating a highway, but we also have the charge from the outer planets using that highway. So we really have 2 fields to deal with, other than the earth and moon.

Unless I'm mistaken, the mainstream view is that the moon hits the Lagrange points L1 and L2, which is where the earth's gravity and sun's gravity are of equal strength (at least L1 is), but that is no where near where the charge fields are of equal strength. If we could use the earth's charge field to corral the incoming charge, then it must have the strength to do so at the distance of the orbiting body.

Actually, it is worse than that. We really need it working behind the orbiting body. The incoming charge must already be in the right trajectory before it comes into contact with the orbiter. That is because the orbiter is part of the gravity mechanism. We need the nuclei to channel the charge and spin it up, etc, etc. That means that the earth's field has to reach past the moon, which also means through the moon, to channel the charge beyond it. Wouldn't the moon break those channels, even if we allowed them to reach that far? Then we need to bring the moon's charge field into it, which would further dissolve the earth's charge channels (and the sun's within the moon's charge-pause).

I was going to throw you a bone to see if you can make use of it when I mentioned the returning charge of the outer planets. As I wrote that, I thought that it might provide a two-way stream that somehow negates each other for our purposes and somehow lets the earth's charge channels act stronger than they are. But I think that would only work, if at all, when the line from earth to moon is orthogonal to the line from earth to sun. Put them in the same line and you lose it. Sorry, no treats today.

Personally, I think binding gravity it too complicated. Not in the sense of being able to understand it, but in the sense of its mechanical processes. It requires so much to happen, and that all has to happen with existing fields that are already being used for various other purposes, and it all has to happen with uniformity. The equations of gravity give you a value that varies with the distance between the bodies. But any gravity mechanism based on charge doesn't have such uniformity. The fields are so much more complicated than that. Again, it might seem to work in a simple 2-body problem, but it fails spectacularly with any more, precisely because it brings in other fields.

The direction of the incoming charge is of great importance. I could allow some-what random angles if we were just talking about charge collisions. In that case, the earth's emission creates the correct line, it collides with some incoming photon that is going towards the earth, but not necessarily straight at it. In that collision the incoming photon might be given the right trajectory, while the outgoing photon moves off based on the angle of incidence. I'm not sure why the incoming photon would do that, and it breaks the spin-up process, but I'll allow it for the sake of argument.

However, that is not the situation we have been given. The collision must happen while moving through a nucleus, and that absolutely requires that the correct trajectories are already present before the incoming charge or the outgoing charge reach the nucleus. It also requires the alignment of that nucleus, so we have 3 things to align before anything happens. That also has to happen in fields that don't have the strength at the distances we need.

I think I need to go over this idea again. I need to spend a bit of time with the papers and see if I can get more out of them. I might be introducing problems that aren't relevant. I just don't know anymore. I have problems with the way I am seeing it, so maybe I need to try seeing it a different way. Use that skill of switching perspectives I talked about. Maybe create some diagrams, they always help, one way or another.

Nevyv wrote:The collision must happen while moving through a nucleus, and that absolutely requires that the correct trajectories are already present before the incoming charge or the outgoing charge reach the nucleus.

Nevyn, if that is the case, than I was wrong all along. But I never understood that part as if alignment / collision must occur while moving through a nucleus. Can you please refer me to the Miles' paper where he explains it accordingly? Or was that your own understanding which is the basis of the above conclusion? I'm asking this because in my opinion it complicates the alignment beyond any feasible solution.

On the other hand, alignment within nucleus presumes the incoming charge field is consisted of mostly charge-recycling photons, i.e. higher order of stacked-spin photons, the ones which are considered as nuclei. That's how I understand your suggestion of alignment happening inside the nucleus. No nucleus, no alignment. But that is an overly complicated process and in my opinion an overhead assumption. Why? The charge field around us is in my opinion consisted of mostly basic energy level photons, since we are able to continuously see everything around us. These are most basic photons varying slightly in their frequency, but still, they account for the observable experience. They get through the dense field without being hit probably because they're to small to get hit. If the incoming field was made of mostly nuclei, the recycling engines with bigger radius, there would be lack of basic photons, which could trigger our visual receptors.

I've already stated that in my opinion the larger bodies emit less energized charge photons, acting as some sort "charge resetting" engines. That is consistent with Miles' opinion, where he says that charge and anti-charge experience considerable de-spin as they cross their paths inside the Earth. That is also consistent with our reality's "request" for the charge field consisting of more basic charge photons - the incredible number of photons recycled each second by individual nuclei suggests that there are many more simpler-to-recycle, basic photons than nuclei.

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I too went to re-read Miles' paper http://milesmathis.com/pause.html, and this time I got  stuck at another issue, vis-à-vis charge field's influence zone limit. Here's what got me thinking:

"The mass of the Sun is 332,990 Earth's and its density is .255 Earth's. We seek a charge density on the surface of the Sun, and we can get that by just looking at the words. We seek a “charge density”. That could be written “charge x density”, and, as I have shown, charge is just a variant definition of mass. Therefore, we re-write the product as “mass x density.” M x D = 84,986. The Sun's charge density is 84,986 times that of the Earth."

What Miles did here is he first translated the mass of the Sun to Earth's terms and then converted that mass to charge density as in equation M x D. In other words, the Sun is not 332990x as massive in terms of Earth's density, but "only" 84986x. He then later compared this translated charge density of Sun to that of the Earth, but in terms of body's own radii - for the Sun he used Sun's radius and Earth's radius for the Earth.

I have two specific questions about this issue. The first concerns surface of the Sun as the charge emitting body, which is much larger than Earth's surface - by cca 12,000x . While the above translate accommodated for mass-to-density ratio relative to Earth, it doesn't account for the surface in relation to density. That's why his result interpretation says only about the general size of charge density: Sun's being 84986x that of the Earth. The statement is completely correct, but only in general terms of Earth's overall charge density.

I think this issue should be addressed accordingly so we can compare Sun's and Earth's density relative to mass and density, but relative also to the size of their surface:

Sun's charge density in Earth's terms = 84986

The calculated charge density 84,986 of Earth's is emitted on much larger surface, meaning its relative density on the surface compared to Earth's is still unknown.

Sun-to-Earth surface ratio = 12,000

To calculate the same relative charge density, we now need to divide Sun's charge density in Earth's terms with the Sun-to-Earth surface ratio:

Sun's charge density corrected by the surface size ratio = 84986 / 12000 = 7.082167

The result means that Sun's charge density on the surface of the Sun is cca 7.082 times that of the Earth's charge density on the surface of the Earth. It came as a big surprise to me considering huge differences in mass and size of both bodies.

The second question is in connection to the first - in the quote above, we can see that Miles has attributed all mass charge potential. In general terms that is completely accurate, but in reality we have both charge as a field (i.e. moving photons of all energy levels ) and charge as a recycling engines, roughly in 19:1 ratio. Not all mass is charge emitted photons capable of transferring energy by collision. The relative charge density calculated by tagging all mass as charge emitting mass should also account for composition of charge emitting bodies - individual elements have different capabilities in terms of recycling charge. On average, they differ by more than statistically significant margin, so in my opinion this should not be ignored.

Airman, thank you for that diagram. As I saw it, I got excited but don't know exactly why. I could see how this may possibly explain the "lock-in" between two bodies. But then I don't know how that would happen.

You already made a conclusion:

"These numbers assume direct radial emissions. Most noteworthy to me is that rather small angular cones at the earth and moon encompass all the earth and moon's direct emissions.

This may be part of why I cannot believe celestial bodies completely align with each other. You can see why the sun can align with all the planets simultaneously, each channel is rather narrow. There's no way to expand that angle, other than say blasting away part of the surface as has occurred on the face of the moon. In my opinion only a small part of the proton matter in each celestial body is charge bound to each other."


Can you please elaborate this further? How are this angular cones related to everything you conclude?

I think it is my mis-understanding, trying to make sense of this paper that seems to contradict itself and every other paper he has written. This is one of the sections that led to it:

Miles Mathis wrote:Now we get to the hard part. Spin mechanics. We take the photon to be a spinning sphere. It can be spinning in any direction, but once it enters a pre-existing field it will be made coherent to that field. How? By collisions. Photons are colliding all the time, and due to their small size they tend to hit edge-to-edge. Edge hits cause spin changes rather than speed changes. Hits can either cause spin-ups or spin-downs. Photons can collide side-to-side, moving in the same direction; or head-on.

When a photon enters a new field of any appreciable density, it will be spun up and down and up and down. When it is spun down, it is weak, and in any hit the weaker particle always takes on the characteristics of the stronger particle (for obvious reasons). If a weak particle spinning on axis-a hits a stronger particle spinning on axis-b, the axis of the weaker particle will move toward b. In this way, over time, the spin axes will be made coherent. It is the same with linear motion, and this isn't just a rule of spin mechanics.

So although photons can be spinning on any axis, we can assume a certain amount of coherence. To simplify the math and mechanics, we then average the field and assign all particles either a left spin or a right spin. The left spin is photon, the right spin is an antiphoton, say. If a photon and antiphoton are moving in the same linear direction and edge-hit, they spin one another down. If they meet head-tohead, they spin one another up. So you have to keep track of spins and linear motions at the same time.

To add to the complexity, most interactions we will be looking at aren't photon-photon collisions. We will be looking at matter fields, so we have to look at how photons interact with matter. As we have seen with the Earth, they are recycled by matter. But they aren't just recycled at the macro-scale, as with a large body like the Earth. They are—at the same time—recycled by protons and neutrons, and thereby atoms. And they also move through matter on two schemes. They move through protons on the pole-to-equator scheme; and they move through neutrons on the pole-to-pole scheme. Since atoms contain both, photons move through the nucleus on both schemes. If the nucleus has a strong carousel level, the main scheme is pole-to-equator. If the nucleus has a weak or non-existent carousel level, the charge also moves pole-to-pole. So this is another factor you have to be aware of. As we saw in my papers on Rayleigh scattering, depending on the elements involved, your spin expectations can be flipped. Nitrogen or Potassium may give you a different field than Tin or Silver, say. That is where we get into conductors versus insulators, but we won't have to include that in gravity. The only way we include it is as above, where the Earth is recycling on both schemes. Pole-to-pole, the Earth is acting as a gigantic conductor. Pole-to-equator, it is acting as a gigantic insulator.

Since I made an analogy to atomic binding in the previous paper, some have thought I meant gravity is a straight analog of the strong force. That isn't what I intended. Although there are similarities, they aren't the same. The nucleus isn't just a little Earth, so while it is good to see the similarities in the mechancial fields, you have to be aware of the differences as well. As I have shown in previous papers, protons and neutrons don't repel one another in the nucleus, so you don't need a force to counteract that. Charge repulsion is caused by photon bombardment upon matter, and although you get that between protons outside the nucleus—where they aren't recycling charge in defined streams—inside the nucleus you don't. There, the photons are kept on proper paths, and they don't keep the particles apart. So in the first instance, the nucleons are bound simply because there is no force to unbind them. Also, they are bound because stars or galactic cores previously bound them with pressure and heat. But there is more to it than that, of course, since they are bound in several other ways. Yes, they are bound by charge pressure from outside, which is a pure sort of push gravity. But more importantly, they are bound by their own charge recycling. You will say that the photons are moving as much out as they are in, which is true. But you are missing the fact that the recycling is always moving in both directions. The nucleus isn't just recycling pole-to-equator, on an in-out scheme with a halfturn. It is recycling from both poles, with charge and anticharge
meeting and crossing. As charge and anticharge meet along the pole, they not only spin each other up, creating current and magnetism, they also create a bond. How? Again, by pressure differences, or field potentials. The same pressure differences that cause the vortices cause the bind, you see. The spin of the proton and nucleus creates a semi-spherical field with polar angular momentum weaknesses. The force in at the poles creates the vortex, and the same force creates the “gravity” or “strong force”. And, as you can see, we can use the same mechanism to create more gravity at the poles of the Earth, despite the fact that the field there is opposite in other ways to the field at the equator.

You will say that in that case, the nucleus would dissolve along the equator. The nucleons on the carousel level should be flung out into space. Yes, we would expect a binding weakness at the equator, one that we cannot make up with straight charge pressure. So the nucleus must have a similar effect at the equator as we will see on the Earth, with opposing photon fields creating another sort of bind. In other words, the charge pressure at the nuclear equator is vastly increased by the spin mechanics at the boundary. The incoming photons of the ambient field are spun-up by the exiting photons of the channeled field, giving them more energy. So when they impact a nucleon, they have more force than they would have, causing a net force in.

You will say that, in that case, when the photons moving along the nuclear pole meet and spin one another up, creating current and magnetism, they should also be energized. In which case they should create a force out. True, except that to create a force out, they have to collide with a nucleon. . . which they do not. Those photons move on down the pole and exit, without hitting a nucleon. That is because they are being channeled. But the ambient field photons coming in at the equator are not being channeled, are they? No, so they are free to collide with a nucleon there, creating a force in. I told you, these problems are complex, but with spin mechanics an answer is always there if you dig deep enough. We have many degrees of freedom that the old theories missed, and all of them are mechanical.

Then this part:

Miles Mathis wrote:Notice that we have a complexity here we didn't have at the quantum level: we have to explain why you are hit by charge coming down but not by charge coming up.

and:

Miles Mathis wrote:And, as with the nuclear equator, the photons moving up spin up the photons moving down, raising their energy. So when they hit you they have more force. You will say the photons moving up also have more force, so they should drive you up. But that is only if they are not channeled, so we must assume they mostly are channeled. And since there are more photons moving up, the photons moving down will be spun up more. All the photons moving down will be spun up, while only some of the photons moving up will be.

I thought that the discussion about spin-ups involved the nucleus, but can't seem to find it now so I assume I connected two parts that shouldn't have been connected quite like that.

The part I took and ran with was this:

Miles Mathis wrote:As charge and anticharge meet along the pole, they not only spin each other up, creating current and magnetism, they also create a bond.

Maybe I held on to the 'along the pole' part a bit too long.

I notice this at the end about orbits, although not related to the above, it reflects on our discussion:

Miles Mathis wrote:What about orbits? Do we have to make changes there as well? Yes, though the math won't change, some of the assignments will change. If we keep gravity as a concept, it will apply only to the unified field binding effect caused by overlapping and interpenetrating charge fields. There is no longer any solo gravity there, either. Mechanically, the centripetal vector isn't caused by mysterious action-at-a-distance or by magically curved fields, it is caused by real charge motions and spin mechanics. In other words, planets inhabit orbits where their incoming and outgoing charge fields balance. They are pushed out by the Sun, and pushed back in by returning charge from outside planets and the galactic core. They are trapped by a complex Solar vortex. Even their sideways motion is explained by the Solar vortex, and not by “innate motion”. There is no innate motion. All motions—in, out, and sideways—are caused by the charge field. But the charge field—although all of a piece—is a veritable honeycomb of influences, from the galaxy, Sun, planets, and moons. These are the only perturbations in the field. Perturbations are never caused by “remaining inequalities” or nonsense like that. Just as in Relativity, math cannot be the cause of motions. All motions are caused by the influence of other
bodies.

Does that mean he is letting go of the repulsion idea? And I still think this falls to my arguments about charge field strength when discussing orbits within orbits. The earth doesn't have the strength at the moon to override the sun. So, by this mechanism, the moon should not be orbiting the earth.

LongtimeAirman wrote:The earth sits, spinning in the sky once every 24 hours, but it never changes its spatial location with respect to the moon.

Except it doesn't do that. The Earth does change its location with respect to the moon - in fact, they orbit a common center of mass (barycenter):


You can see a simple video of that here, though I couldn't embed it:
https://www.av8n.com/physics/earth-moon-spin-orbit.htm

And this is true of every orbiting body, perturbed by their centers of mass. The Earth orbits the sun-Earth center of mass (which is relatively near the center of the sun, but not AT the center), and this orbit is perturbed by all other planets and bodies as well, even if only slightly in most cases. AND the orbits are perturbed by classical Miles charge, the kind that causes elliptical orbits and gives us that balancing mechanism.

I don't think this is unimportant, but rather critical to charge-bonding "gravity" theory. It compounds Nevyn's chief issue - how can the MOON "pull" the Earth that far away from itself, when its charge is being trumped by the Earth's AND the sun's, at that distance?

The moon is "pulling" the Earth nearly the moon's own radius away from itself.

Vexman wrote. Can you please elaborate this further? How are this angular cones related to everything you conclude?



Thanks for asking Vexman. Here are two more variants. By now, you should be getting sleepy, … .

I think of them as charge channel diagrams. The diagram is intended to show two overlapping emission fields of two charged particles or celestial spheres, represented by the red and blue circles. It turns out that the field lines between the two objects is interesting. The emissions are shown as outward radial lines in all directions - either top or side views apply. All the radial lines tangent to the other body describe opposing cones. One can vary the diagram by changing the size of a body; or indicate greater emission density with more lines.

The direct emissions within the two conic volumes of space between the two bodies are delineated. The bottom image omits all non-pertinent emission lines. Note that all direct emissions received by one body must have originated in the small slice of conic section of the other body. The center diamond, or 3d spindle area is mainly filled with mostly head-to-head opposing photons, horizontally passing each other results in the least amount of collisions.

Direct photons traveling off of the center line must encounter an increased possibility of collision with the additional radial lines of ever increasing angle (with respect to the incident emission) shown outside of each cone – the green and magenta radials. These lines are conducive to positive deflections, a possible mechanism for “concentrating” deflections.

As I said previously, in my opinion, the only protonic charge binding that can occur must involve these two conic sections.

Your feed back is appreciated.  

Thanks for the additional info Jared, I believe my windows 10 won't allow me to open your link. All I was describing was the constant moon/earth link; that the moon always keeps it's face aligned to the direction of the Earth.
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LongtimeAirman wrote:Thanks for the additional info Jared, I believe my windows 10 won't allow me to open your link. All I was describing was the constant moon/earth link; that the moon always keeps it's face aligned to the direction of the Earth.

Sure, I know the moon is tidally locked. But the point remains - the moon is "pulling" on the Earth in the same way that the Earth is "pulling" on the moon. So how does the moon align to the Earth, but also to the sun, and also to every other body in the system? How does the Earth do the same thing? For charge-binding gravity to work, there must be alignment of the primary charge emission fields - in this case, the Earth's must align the moon. The moon's charge emission cannot also align the Earth at the same time it's being trumped by the Earth's charge alignment - nor as Nevyn said, align itself to every other body in the universe as well. Even if we're talking about apparent "pulls" at the nuclear level, the moon would be ripped apart - and yet it doesn't rip apart. In fact, its own gravity holds it together. In which case "charge binding gravity" becomes circular. It's the cause of itself.

Understanding Nevyn's logic and points prior to this conversation, since we've all talked about it considerably, I can't help but agree with him. I'm adding points (or half-points, depending) as I find them to bolster his main point, but like I said more analysis won't hurt. I'm not giving up so much as trusting my guy on this one. Miles may yet drop something on us that addresses these points but thus far that is not the case. Not his problem, again, it's mine. I don't mind disagreeing with him on a particular topic any more than he would mind. It's important that we're able to disagree in the first place. To be skeptical. But it's also not his job to convince me personally, so here we are hashing it out still. Let's continue.

LongtimeAirman wrote:1. Protons cannot be represented at the charge scale as a single b-photons. Blindly sticking to the stacked spin as a single b-photon model omits the presence of 95% of the known proton’s mass, the charge it is recycling at any given moment. The stacked spin single b-photon model may have the correct field pole profile, but by itself, it cannot explain the proton’s disc-like emission profile. The stacked spin single b-photon model needs to be expanded to include charge.

I disagree. They both can be represented at the charge scale as a single photon and have been represented this way, in all of my sims and all of Nevyn's apps. Nobody here is "blindly sticking" to the model, we are diagramming and illuminating it, visibly and visually. I have never omitted the charge field in my nucleon videos, which in fact show primarily the the charge field's motion through the nucleons:

https://vimeo.com/320694284


And another one, this one where the nucleons are held apart ONLY by charge photons, with no keyframes or animation techniques:

https://vimeo.com/208245448


So your first two sentences are false. Your third one seems to say a single B-photon cannot create the disc-like charge emission we see in the diagrams, but then Nevyn already addressed that too and you've seen it yourself:




Turn that second one on its side. That's the path a proton's B-photon takes, as we understand it. Miles has explained this in many papers and Nevyn just hit it again a page back in this very thread:

Nevyn wrote:
With respect to charge vortices, I don't believe that they are vortices, but they might look like one. This is just the result of the way a charged particle pushes charge around. It prefers the equator to emit charge and prefers not to emit it at the poles. It still can and does emit at the poles, but the majority of charge is pushed to the equator. That is because the equator is in the same plane that the charged particle's top spin level is moving in. If the top spin level is a Z spin, then the motion is in the XY plane. Therefore, the equator will be in the XY plane too.

This creates a repulsion at the equator that limits incoming charge from this area. Therefore, the poles still receive the same charge they did without the particle being charged and it looks like it is pulling charge into itself. That is an abstraction, though, for it can not pull in charge. It just accepts what is already going that way. But the equator does not just accept anything moving its way. It is a bully, and tries to get rid of it before it can actually strike the charged particle itself.

And thus your last sentence in that quote I just quoted is also false - the model not only HAS been expanded to include charge, it ALWAYS HAD included charge. From the very beginning, when Miles first explained it to us. Polemics matter in these things, and I feel like when all of your statements are false like that there's got to be some reason.

So is what you're proposing then that, as you've said, a great many B-photons would be either following the main B-photon around the proton's path? Or that a proton is a great MANY of these high-mass, large-radius B-photons at that spin level? Because then you have not only the proton and its 19x mass in charge photons, but you said perhaps thousands of MORE protons, which is a huge mass surplus. You're proposing that the proton's mass is not just the proton PLUS its charge, but the proton and a thousand more protons, effectively.

And also what mechanism would cause these unwitnessed ghost-protons to trail the original one? We can't have gravity at this quantum level anymore in the charge-binding theory, so you can't use that. What would possibly cause all these thousands of ghost protons to follow that path? And WHILE they are following the path, it is also MOVING through space too - even a proton here on Earth is. Everything is moving, relative to something else or some rest position. So how could these unseen, undetected, over-weight ghost-protons know how to follow our orginal's path?

And how would we not measure the bloated ghost-protons as well, making the proton thousands of times more massive in any and every experiment?

Vexman, I don't think I am understanding your issue with charge density. In that math, I believe Miles is finding the total charge density of the bodies. He is using the volume, so it is total charge generated by all of the mass inside of that volume. That charge is emitted at the surface, but it still has that density because all of the mass is emitting. Of course, this is averaging the charge over that surface, and there can still be polar and equatorial differences that are ignored in this problem.

I'm not sure why you are bringing the surface area difference into it. I don't see how that could impact the problem. The sun emits over its own surface area, not the earth's. The reason Miles is converting from sun-units into earth-units is so that we can compare the fields. That is why the charge-pause is specified in earth radii. It is a relative value, rather than an absolute one.

Maybe you could explain it again for me, and I might see what you are getting at.

I think I may have done something similar to what you are describing. At least, I discussed the idea when I thought I had one of those Aha moments. Have a read of this and see if it helps you in any way.

https://milesmathis.forumotion.com/t476p125-possible-charged-particle-field#4174

You need to follow that on for a few posts, even onto the next page, because I do find an error in my initial analysis that I correct a bit later. It uses expansion, but the way I am using it might spark some idea for you.

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Jared wrote. Sure, I know the moon is tidally locked. But the point remains - the moon is "pulling" on the Earth in the same way that the Earth is "pulling" on the moon. So how does the moon align to the Earth, but also to the sun, and also to every other body in the system? How does the Earth do the same thing? For charge-binding gravity to work, there must be alignment of the primary charge emission fields - in this case, the Earth's must align the moon. The moon's charge emission cannot also align the Earth at the same time it's being trumped by the Earth's charge alignment - nor as Nevyn said, align itself to every other body in the universe as well. Even if we're talking about apparent "pulls" at the nuclear level, the moon would be ripped apart - and yet it doesn't rip apart. In fact, its own gravity holds it together. In which case "charge binding gravity" becomes circular. It's the cause of itself.

Airman. I don’t know, I can't explain, we're trying to work it out. My core belief is the charge field. My particular belief that all nearby celestial bodies in a system are linked. Obviously, Miles hasn’t made charge binding very clear yet. As I recall, you started this thread because you needed help in understanding charge gravity. It’s been a very difficult/rewarding discussion. I thought you made up your mind, as well as your comments here make clear – charge as gravity is unlikely; and allowed that we could continue our discussion. A review of which would make it plain that nobody seems to know how charge binding works.

We know the sun is linked to all the planets. How does that work? The charge channel model I posted four images of provides the justification for what I believe is the main idea – celestial charge links can be described by radial emission cones from the center of one object to the tangent of the other object. I belief all the protonic alignment involves less than a few angular degrees - as seen from the sun. As such, they usually all fit on all bodies, at all times without crossing or overlap most of the time.

Jared wrote. Understanding Nevyn's logic and points prior to this conversation, since we've all talked about it considerably, I can't help but agree with him. I'm adding points (or half-points, depending) as I find them to bolster his main point, but like I said more analysis won't hurt. I'm not giving up so much as trusting my guy on this one. Miles may yet drop something on us that addresses these points but thus far that is not the case. Not his problem, again, it's mine. I don't mind disagreeing with him on a particular topic any more than he would mind. It's important that we're able to disagree in the first place. To be skeptical. But it's also not his job to convince me personally, so here we are hashing it out still. Let's continue.

Airman. You agree with Nevyn, but please let us disagree. We need to figure out how whether or not charge binding does or doesn't work. You seem to have your mind set. I understand, Nevyn and you agree that the so called alignment must involve the entire object, I happen to disagree, and I've explained why.
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LongtimeAirman wrote:You agree with Nevyn, but please let us disagree. We need to figure out how whether or not charge binding does or doesn't work.

Definitely. By all means, we need to disagree and counter each other (or not) to learn anything here. Miles has done us all kinds of favors and given us many tools but it's not his job to come here and hash shit out in his free time. It's our voluntary task and I enjoy it - and I hope my previous comments weren't meant to seem stifling! By all means, discuss.

LongtimeAirman wrote:We know the sun is linked to all the planets. How does that work? The charge channel model I posted four images of provides the justification for what I believe is the main idea – celestial charge links can be described by radial emission cones from the center of one object to the tangent of the other object. I belief all the protonic alignment involves less than a few angular degrees - as seen from the sun. As such, they usually all fit on all bodies, at all times without crossing or overlap most of the time.

See, I completely agree with this assessment! And your cones work for me, for head-on charge definitely. And to illustrate the cross-sections involved. I'll try to isolate those in a diagram too, using my planetary tilt scene, though as it stands the actual (relative to say, the Earth, if we used its radius as "1") distances would severely diminish those angles of course:



It's not surprising but interesting that the first four planets have no real charge overlap. That is, they don't block each other from solar charge ever in their orbits (except if they cross but I haven't gone into that in this model, this is just their tilt and eccentricity). But the Jovians all overlap.

I don't quite know how that helps us with charge-binding gravity so I'm studying your diagrams and explanations above to see if I can figure out how that would cause a binding. Or how it could/would address Nevyn's point about alignments, if possible. I'm not TRYING to have a negative bias here, but just like with the higher axial spins we discussed before I simply don't see how some things could work. Doesn't mean they don't, just means I'm either slow and wrong or slow and or right. Probably slow, in either scenario!

Before you two go too far, I want to point out that I have admitted to a mistake in relation to needing the nuclei to align. I think the charge still needs to align, although it might relax that requirement a bit too. I don't know at this stage. I need to re-re-re-read it and try again. We will also see what Vexman can come up with, without me hounding him with my own problems. At this stage, I don't think any of us have a good model. I'm not even sure Miles does. I think Miles is using broad strokes to paint this picture, but I want him to use a finer brush. I need the details.

I still think there are issues and some parts seem a bit too selective to me, for example, charge moving up channels but charge moving down collides, just because you are aligned to the earth's field and not the sun (close to the surface of the earth). Well, there is at least 1 moment of every day where the earth and sun and you are all in-line. Shouldn't gravity disappear at that time? But I need to give it more time and effort.

It's been a great discussion so far and I think we've all progressed from the starting point. At least I did, so thank you all for that. I don't have all the answers, far from it. But I like this situation better than with the expansion theory of gravity, which too has left me with more questions than answers. That's just my subjective opinion, not worth much in view of real physical arguments and mechanics.

Anyway, the point with my attempt at calculating relative charge density with respect to surface area is to show that Sun's charge field isn't that much stronger at its surface in comparison to the Earth. Yes, it has roughly 85000x Earth's overall charge potential due to much larger mass, but than it emits it from 12000x larger surface. So in relative terms, we can't compare it 1:1. I can understand the comparison between overall charge density and I have no real argument against what Miles did with his math. I'm only wondering if distributing all mass as charge over the entire body's volume is completely rational when we're looking at a rather minuscule part of it acting upon other bodies.

On the other hand, when I was observing Airman's diagram, I thought that maybe we're not interested in 100% of body's charge potential. Looking at Sun, we're receiving only the proportional volume of Sun's emissions, which correspond to charge coming in from a relatively small radial arc of the whole Sun's sphere. For each Earth's radial channel, there are 7 Sun's radial channels. Considering the huge difference in sizes, arc curvatures and distance between bodies, I don't think all 7 channels are able to aim at / reach Earth. The distance between the Earth and the Sun is roughly 214,5 Sun's radii, which means the channels have substantial separation between them - the conical empty space between them is much larger considering the distance and it grows bigger with distance. Now, following this line of thought, the Sun's charge density which would account for only small arc of Sun's surface, is possibly the reason why Earth's charge field can reach the Moon and create the binding force needed for the lock-in. In the same context, Sun's field affecting the Moon comes from even smaller arc of Sun's surface, while most of Sun's emitted charge never reaches or influences the Moon. In essence, both Earth or Moon have larger arc of emmiting surface pointing to each other, than any of them has towards the Sun or vice versa. So in my opinion, the proportion of Earth's or Moon's surface arc emitting charge is not equal to the proportion of Sun's emitting arc. Smaller bodies will have larger surface arc emitting charge in comparison to the bigger bodies when facing each other, which can be clearly noticed from Airman's above diagram. But then Sun's charge density is much bigger, so does that have any influence in this? Can this help in a) establishing the points of  body's charge field influence zone limits and b) calculating the strength of each body's charge field relative to the surface of emitting body, incorporating the distance as well. All this was in my opinion in connection to the dilemma about the reach of Earth's charge field and mechanism of Moon's orbit lock-in to Earth instead of Sun.

The two points which absolutely need to be resolved are alignment of opposed charge directions of movement which enables spin-up, and presence of charge field of each individual body and each particular field's limits (at least understanding what sets those limits).

I may be completely wrong about this all. It may be a dead-end and leads to nothing but a waste of our time. It was just another idea that was in my thoughts and I shared it here, as it may have sparked some other ideas. Or it may be "trashed" as irrelevant. I don't really know about the correct answer, but I want to believe we're moving forward either way.

LongtimeAirman wrote:Direct photons traveling off of the center line must encounter an increased possibility of collision with the additional radial lines of ever increasing angle (with respect to the incident emission) shown outside of each cone – the green and magenta radials. These lines are conducive to positive deflections, a possible mechanism for “concentrating” deflections.

As I said previously, in my opinion, the only protonic charge binding that can occur must involve these two conic sections.

I have tagged the radial lines on both bodies just to make my point. I think that the intersections might be a mechanism not only for concentrating deflections, but also for the spin-up. The requirement here is the correct spin of colliding photons and the correct spot of impact.

Also, in the view of possibilities for right kind of impact I think the occurring protonic charge binding includes neighbouring conic sections too, not just the central one.

The increased possibility of collision with the additional radial lines is true, but it doesn't exclude a possibility of "cross-radial" photon collision. For instance, a photon coming from magenta #1 line (M1) could miss a photon coming from green #1 line (G1), and collide with G2 (or even G3). If G2 (G3) would collide with M1 at its bottom edge or far right equatorial edge with the correct spin, it would align it towards the source of green photons, spinning it up at the same time. The redirection would align M1 photon toward an area where G2 (G3) originated from, which means that the area of protonic charge binding involves G2 (G3) section too. But certainly the limit is at the N-S pole, with less and less probability of photons in "cross-radial" collisions. By diminishing probability of such collisions as we move from the central line, the vast majority of affected area would be a few radial lines away from the central line.

Vexman wrote:So in my opinion, the proportion of Earth's or Moon's surface arc emitting charge is not equal to the proportion of Sun's emitting arc.

I absolutely agree with this, and with LTA's take on it so far as well. My diagram is accurate in relative SIZES though not distances, so the cones or arc of charge would actually be much SMALLER than I drew it, but it shows us the same thing LTA was trying to outline. I'm still not sure how that helps us though, since as Nevyn has pointed out the Earth's charge-pause also exists, which is a boundary where the sun's charge has more density even if it's hitting the moon with only a tiny cross-section of its total emission. Thus, the magnetopause, the "coma", and the solar wind.

The part I just can't power through is how charge can be repulsive and attractive at the same time. I know that's too simple, and Miles has stated pool-ball mechanics aren't enough here, but I don't understand the fields involved or how they would "trump" bombardment/repulsion yet. Re-re-re-reading it all as well.

Jared wrote:I'm still not sure how that helps us though, since as Nevyn has pointed out the Earth's charge-pause also exists, which is a boundary where the sun's charge has more density even if it's hitting the moon with only a tiny cross-section of its total emission. Thus, the magnetopause, the "coma", and the solar wind.

OK, so we can end this debate about whose field is dominant at the Moon :

Miles Mathis wrote:The ambient field (the field around the Moon) isn't upside down relative to the Moon, but it is opposite in spin in another way. Since the Moon is so close to the Earth, the Moon's ambient field is determined more by the Earth than the Sun. So the field emitted by the Moon is always meeting the field emitted by the Earth head-on (at least on the nearside). Since the linear vectors are opposite, we again get a spin cancellation.

The Earth's photons only set up a sort of wall, keeping the Sun's photons from defining the ambient field, and therefore providing the boost themselves. The Moon's photons on the farside are emitted into a flat field, as it were.

http://milesmathis.com/marsmag.pdf

I believe this leaves no doubt about the dominant charge field at the Moon, even though magnetopause exists at the same time. Like I said, magnetopause is about the solar wind, carried by charge field but made mostly of ionized content, meeting Earth's field head-on. Solar wind is going at .133  percent of c in average, so it's not exactly typical charge field. It doesn't say anything about the field's limits and strength and apparently Earth's field can be dominant at the Moon's distance. Since Miles has left out the reasons why that is so, one assumption is related to emitting body's arc size.

Why is all this important? The binding theory fails spectacularly if Earth's field isn't able to reach the Moon and fails to cause a spin-up and lock-in to Earth. This should prove such assumption wrong.

The conical shaped channels of incoming charge could explain both the alignment and spin-up process as I already wrote about earlier. So what I think is happening is that we're discussing the options for all this to come together, tossing around ideas and thinking painfully hard while at it.

I see some problems. Firstly, Miles is calculating the charge density. A density is the amount of substance in a unit volume. It does not represent the total amount of substance on the surface at any given time. To get that total you would have to multiply by the surface area. So you can't divide by the surface area from the density to limit the amount of substance to that which is aimed at the earth, for example. What you should do is calculate the area of the earth that is projected back onto the sun and then multiply that number by the density of charge at the sun's surface. That would give you the amount of charge that reaches the earth, or some number related to it. But we don't need to do that because we can just work in densities, since they already have the concept of unit volume. As long as we are using the same units for both bodies, then we can compare them.

Secondly, you are all treating these lines as if they are physical lines of charge. They are not. At best, they represent the density in the same way that magnetic field lines represent it. They are an abstraction to make it easier to visualize a field.

Thirdly, the diagrams should be using the charge density of each body to determine the angle between the radial lines for each body so that they can be compared.

To do that we take the charge density of each body, C = MD, and divide one by the other to get the ratio of their charge densities at their surfaces. That ratio is then used to determine the angles between the field lines on each body.

Cs = Ms * Ds = 2.804e+42 kg/km^3
Ce = Me * De = 3.304e+37 kg/km^3
Rse = Cs/Ce = 2.804e+42/3.304e+37 = 84867

Therefore, if the angle between field lines on the sun is 1°, then the earth's field lines would need to be 84867° apart. Which isn't a very helpful angle, so we can work it the other way and start with the earth. Let's give the earth an angle of 10°, therefore the sun's angle would need to be 10/84867 = 1.1783e-4°. Note the I have simplified the numbers in this text, but used more precise numbers in the calculations. I used the numbers from my Solar System page, so they will differ from Miles numbers a little bit.

So you can already see that the sun's charge field is more dense than the earth's, just from those angles. Five orders of magnitude of difference. That is why the earth's field has no chance against the sun's. The earth is at 23455 earth radii from the sun, but its charge field only reaches 11 of them with a density to overpower the sun's density. That is 1/2132 = 0.0004 of the distance.

Note that because we are using angles to create radial lines out from the surface of a sphere, the curvature differences between the bodies are taken care of too. The earth not only has a lot less charge to work with, but it also has more curvature, so its field will diminish quicker using this field line method. Another way to say that is that the channels will dissolve quicker.

I'm not sure if that quote from Miles helps or not.

He states that the moon's field meets the earth's field head-on. But it also meets the sun's field head-on too. Just imagine yourself standing on the moon. Can you look directly at the sun? Of course you can, at least at some times. Most times.

Then he states that because the fields meet head-on, we get spin cancellations. Which directly contradicts the binding gravity theory where he uses head-on collisions to cause spin-ups. Which is it? Whatever he wants in the moment?

Most papers Miles states that photonic fields don't collide with each other (much), often quoting evidence of that from mainstream sources, but in binding gravity, they suddenly collide all the time. Which is it?

He shows us the calculations of the magneto-pause, explains why it is where it is and why it relies on the charge field. Then extends that field further, much further, with no explanation in another paper. Which is it? I don't think that is a matter of looking at a different problem in a different way, but I haven't read that paper recently, so I leave open that possibility.

I don't think that quote solves anything, other than using an appeal to authority, which is baseless and Miles would agree, and has stated so many times. As you point out, it doesn't even explain why, it just makes a statement. He needed it for that problem and used it. Maybe there are justifications for it, I don't know, but that is what we would need to solve our problems. We need the substance, not the statements.

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Hey Jared, seeing the confusion, good you caught your error, baring my own mistakes, this chart shows the relative sizes of the Sun and planets. From the sun, Venus appears to be the largest planet.


Nevyn wrote. I still think there are issues and some parts seem a bit too selective to me, for example, charge moving up channels but charge moving down collides, just because you are aligned to the earth's field and not the sun (close to the surface of the earth).
Airman. I agree, I see plenty of confusion too, but I'm used to that. This diagram with earth/moon details including radials provides a example of a comparison/difference between out/in charge collisions/channels because of different field curvatures.

The photons originating on the moon and headed for the earth have a much broader target; moon emitted photons cross earth radials over a longer distance which results in more deflection, and concentrating capability of the larger earth. Earth photons headed to the moon are much closer to the moon when they pass by the moon radials for a shorter distance and time. It is easier for the earth radials to redirect incoming moon photons than for the moon radials to redirect earth photons.

Vexman, the diagram always seemed good to me, after you started describing it, even better. I realized that – of course - the center line between the two objects is a spin up trajectory. And the concentric radial zones do add to the positive deflections.

P.S. Oh, the photons between earth and moon don't spin-up they cancel, that's just wonderful. And another thing, best discussion ever.

P.P.S. Whoops, I take that back. I see that Jupiter looks bigger than Venus.
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LongTimeAirman wrote:Vexman, the diagram always seemed good to me, after you started describing it, even better. I realized that – of course - the center line between the two objects is a spin up trajectory. And the concentric radial zones do add to the positive deflections.

P.S. Oh, the photons between earth and moon don't spin-up they cancel, that's just wonderful. And another thing, best discussion ever.

I think that's another one of my problems when we talk about spin-ups and spin-downs, or rather when Miles does. We have no reason to think that a head-on vector has any greater chance to cause a spin-up or a spin-down, necessarily. We can run the math on something like that (Nevyn can, anyway!) so that may become very important here.

So far in my study, spin-ups and spin-downs are completely dependent on two things, not just one:
1. The vector of motion relative to the A1 spin
2. The vector of motion relative to the top spin, which is a compound motion (acceleration, or "curve")

This is another way of expressing "c²" in the energy equation basically. So it becomes something like c_sub1(axial-spin-vector)*c_sub2*(top-spin vector)*c(linear), though I don't think that's quite right. Point being, not even close to ANY collision will cause a spin-up or spin-down. Only very, very specific collisions cause spin changes. The most uncommon collision causes a dead stop. But these spin-change collisions are only slightly more common relative to regular, bounce-out spins that simply redirect the existing photons with no spin changes. Those in my opinion are vastly more common, and it seems like we can verify this since most photons still go light speed. If dead-stop collisions were any more common, we'd see various diminishings in the field.

For the most part his theory is damn great, and has a LOT of accuracy (matches data). In many ways and theories I think he's dead-on. Just like you, we've gained a lot so far from him and it's a few finer points (for me, axial spins, for you, proton/charged particle activity, for example) that we're still studying heavily and working on - sans Miles.

This isn't a nebulous point. Miles has given me little or no commentary on my most important simulations - and that's not in any way a falsification of his theory, it just means he hasn't studied the collisions up close yet other than perhaps to watch them once or twice. Which is fine. But the thing is he CAN'T study those collisions up close without similar diagramming. It's not even possible to do, inside our heads, or everyone would just do that and physics would have never been mysterious.

Now this isn't to say my sims are important because they're MINE, or that they're important because they're ACCURATE as fuck, because they may not be. But between Nevyn's apps which are PRECISE as fuck and my video-diagrams which are as accurate as I can make them, we have two tools he doesn't and didn't have. So we are able to study these collisions in ways he cannot, nor can anyone else.

This doesn't mean my vids or analysis is the only correct one here and I hope that's clear. It simply means we CAN study these collisions and, with a bit of math and then a bit of diagramming, determine the propensity for spin-ups and spin-downs given two or more field densities and field VECTORS. The sun/Earth/moon problem is the perfect example.

But again, all that is really interesting but it doesn't yet help us with charge-binding gravity at all. It only allows us to determine the propensity for spin-ups and spin-downs based on vector and angle, which really does nothing for the theory either way. It doesn't give us a vector DOWN when we have a massive vector UP to begin with. In charge-binding gravity we don't have the collision to work with, it seems like - since charge-binding is now given as the OPPOSITE of a collision, but an attraction. A collision isn't an attraction.

I imagine you guys are sick of hearing this by now but I actually have the same problems with this that Nevyn does, again. Also. This isn't a matter of politics but a matter of logic to me. While this theory evolves and we dissect it, these contradictions need to be addressed. We've been given a colliding, repulsive particle and field and now it's been flipped to be attractive at some times yet still repulsive at others.

I honestly feel like Miles simply isn't explaining it very well yet. I think he can do better.

Just thinking aloud here, trying to track a single individual photon being emitted from the Earth. An emitted photon cannot have any collision or transfer of energy back down towards the Earth unless:

1. It bounces another photon down at the Earth or
2. It bounces back down at the Earth, itself

Any other path, collision, or transfer of energy, be it through-charge or magnetic (parallel to the Earth, by being emitted perpendicular to any matter it recycles through) would not be a bounce "down". No downward vector would be involved except in those two cases, it seems to me. None of that charge COULD hit the moon anyway except through-charge, since all the other cases involve changing the vector and thus not emitting anywhere near the angles necessary to hit the moon, relativity or none.

So is Miles saying that because these emissive photons don't go straight up, being deflected by matter itself either in the Earth or in the atmosphere, that creates a sort of gap in the field where outside photons can then create the pressure DOWN that accounts for gravity? Is he saying the gap in the field created this way causes the ambient field to attain the 9.8m/s², whereas the Earth's field is tamped down to that .01m/s² or whatever?