The Cause of Gravity - the next major chapter

I don't think that is what Miles is saying. He has swapped the vectors around, so the earth is emitting the 9.81m/s^2 and the charge coming down creates the 0.009545m/s^2. That then means there is a greater vector up, not down as previously assumed. So we have to explain how a great vector up doesn't push us away like it should.

To solve that, he is using binding. Basically, as I understand it, when the nuclei of the gravitating body (yes, I keep coming back to this, even though I thought I was wrong about it, and I was in one way) are aligned with the earth's field, then they channel most of that charge instead of being struck by it. Therefore, they do not impart a velocity to it, or at least not much.

My previous understanding was that the nuclei were the conduit for both charge fields and also the cause of the spin-up in some way. Now, I believe that the nuclei must align in order for the upwards charge to be channeled instead of causing a collision. Apparently the downwards charge doesn't get to channel and always collides (or a majority of the time).

How does that change things? Well, for one, the nuclei still need to align to the source of gravity (for want of a better way of saying that), but the incoming charge doesn't need to align with it too. That means the bodies charge field can cause that alignment, as Vexman has been trying to do. It also means the incoming charge can be random, which gets rid of some of my arguments, and gives a reason why it would collide more than channel.

However, I believe that my main argument holds. At the distance of the moon, the earth does not have enough power to align the moons nuclei to it. Vexman did find a corroborating quote for the alignment, but lacking a mechanism, I find it lacking power. Well, it has a mechanism, just not a very convincing one. It is never explained how the earth overpowers the sun, at the moon. Can we imagine some of earth's charge reaching the moon? Sure, but why can't we also imagine that the sun's charge does too? What excludes the sun's charge? The only answer is because the earth-moon charge is head-on to each other. As if the sun-moon charge isn't. Maybe we could use curvature to help explain the difference, but Miles didn't in that paper, and I would want less curvature rather than more, but there may be a way.

How do the spin-ups come into it? That does require an alignment of both charge streams, if the photons need to create a vector radial to the earth. But maybe that is a different part of the paper, not really about the gravity vector itself. I can't remember right now. I'll have to look into that.

Something I found nagging at me while reading these gravity papers, but couldn't really understand if it was right or not, is that Miles appears to be talking about objects that are on the surface of the body, rather than orbits. I am wondering if he is saying that gravity doesn't have anything to do with orbits, and they are just the path where all forces are equal for a given orbiter. That would change a lot of what we have been talking about, and the problems we have. If we don't need the moon to be aligned to the earth, then my charge density argument falls in a heap.

What do you all think? Is that getting closer? Is Miles talking about surface interactions and not orbits?

Nevyn wrote:Now, I believe that the nuclei must align in order for the upwards charge to be channeled instead of causing a collision. Apparently the downwards charge doesn't get to channel and always collides (or a majority of the time).

How can any charge be "channeled" without a collision?

At the Earth's surface, which is rough and not remotely spherical (for example) at any instance except as a whole, charge is emitting in many directions - but chiefly a relative "straight up" simply due to the density and radius of the planet. There's enough matter below any given square meter (say) of surface to be emitting enough straight-up charge photons (from whatever source vector) to "dodge" any matter above it, as a total. But even if we give the surface itself a total straight-up vector, we still have MOST photons moving straight up. That's what emission is to begin with, yes?

So any matter above the surface (which would of course be the atmosphere) can only "channel" the emissive charge field by direct collision, either with the actual B-photons themselves or the B-photons they are channeling, yes?

To paraphrase your frustration, if the downward "ambient" and solar photons must collide with matter to create any field effects, the upward "Earth" photons must also collide with matter to do the same. Without any collisions there's no transfer of energy, right?

There's no actual mechanism for a channeling without a collision, it seems to me. What could "channel" a field without somehow deflecting/reflecting/refracting it, since any change in a photon's direction must be due to a collision of some sort? Since we've discarded all spooky forces and actions, the collision is the only one - sans expansion obviously, which gave us a much simpler math at the very least.

It seems quoting Miles was more than an appeal to authority. It directly and undoubtedly negates the point about charge binding in Moon-Earth case, while it confirms Earth's field presence beyond magnetopause or 11th ER. Go figure. So for what it concerns me, I'm back to the starting point, scratching my head again. While at it, I'm returning the question back to you: which one is it? Do we have a charge field extending beyond the magnetopause or not? In the case of Venus with 0 magnetosphere, we have it, while in the case of Earth, we don't have it. We can't seem to avoid the results of charge field density calculation, which shows there's actually a point between opposing fields, where we have a charge density equilibrium as we compare them. You can't even imagine how many times I tried to work that over, hoping I could find something to correct it.

Well, uhm. . .  that's a complete show-stopper for me, a spielverderber in the local slang.  

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On the issue of charge density in relation to the size of emitting surface, I have to disagree. But it's only for the sake of the argument, because I no longer think this issue is relevant to charge binding per se.

Your calculation:

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

. . . is equal to Miles' own calculation ; 84867 vs 84986 in end-result, respectively, in terms of Earth's charge density. You took the mass of each body and multiplied it with body's own density, than compared them to get the relative charge density measured in Earth's units. It interprets like: if Earth has MxD equal to 1, then Sun has MxD equal to 84867x that of the Earth. This still says nothing about the surface area of the two bodies, all it explains is mass-to-density relationship, C=MxD, in Earth units. While I completely agree here about the relative overall charge density, charge density at the surface of each analyzed body is not yet set as relative.

I think the correct relativization of charge density includes the surface area translated to Earth's units, as said before:

Sun's charge density over volume: 84900 <-> Earth's: 1
Sun's surface: 12000x of the Earth's <-> Earth's: 1
ratio density/surface for the Sun (in Earth surface units): 84900/12000= cca 7
ratio density/surface for the Earth: 1/1=1

What it means is in my opinion this: one part of Sun's Earth's surface equals to 7 parts of Earth's Sun's surface in terms of Earth's charge density at its surface. To properly interpret, we need units to measure the surface charge density, i.e. number of photons / cm2, or similar. If Earth has a value of 1, Sun has 7. This may be completely irrelevant, unless one would want to draw a diagram with correct representation of radial lines. Or calculate the charge field strength relative to the arc of emitting body, which may be related to finding the individual charge field limits. But even that requires proper definition of field presence, measured in so far non-attributed units.

The solution to this mess is so easy, though. Can't believe nobody here thought of it before.

Somebody please call a messenger photon. He'd simply whisper to all those photons spinning the wrong way to either spin-up or they'll get kicked out into the 12th dimension.   Does the trick every time.

Let's keep it simple, guys!

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The new thread page number nine seemed lacking without a channel diagram, so, voila.


Nevyn wrote. It is never explained how the earth overpowers the sun, at the moon. Can we imagine some of earth's charge reaching the moon? Sure, but why can't we also imagine that the sun's charge does too? What excludes the sun's charge? The only answer is because the earth-moon charge is head-on to each other. As if the sun-moon charge isn't. Maybe we could use curvature to help explain the difference, but Miles didn't in that paper, and I would want less curvature rather than more, but there may be a way.
Airman. It's never been explained? Sorry to be so repetitive, I believe that between the Sun, Earth and Moon, all links coexist at all times. I added the moon and a plot to show the data. The moon’s angular width, seen from the sun, is (of course) 3.669 times narrower than the width of the earth. The sun delivers somewhat less direct emissions to the moon. The earth/moon link is too strong to be overpowered by the sun. The moon is also spinning with respect to the sun; but not with respect to the earth. I believe the earth/stationary moon link is inherently stronger then the sun/ spinning moon link.
 
The field lines show that there is always a head-to-head natural direct emission pathway between two bodies – the strength of which is a function involving the addition of both the earth and moon. I’m under the impression that the direct channel between the two object has charge field properties that reinforce or somehow multiply the earth/moon charge link. In other words, the earth/moon link is always passing a two-way charge current greater than either the earth or solar charge fields (at the earth’s orbital radius) alone. The sun is too weak to overpower the direct earth/moon link. However, both the moon and earth also have direct links to the sun. I plotted the earth/moon links in the 12, 1, 2, and 3 o’clock positions of the solar orbit but don't see any value in adding it, the sun's radials at the earth/moon are essentially parallel lines back to the center of the sun.

Where did the 7 channels for every one of earth’s come from? If the sun is emitting 7 channels for every one of earth’s, the great volume the sun is emitting into is more than seven times larger at the earth/moon to sun orbital distance. Anyway, curvature is extremely important, and necessary in properly identifying relative field strengths. I agree with Vexman, we don’t have a proper means to compare our fields. Without the proper field density strengths, the radials in these diagrams just indicate direction and the inverse square law.

Please feel free to make requests or suggestions. How can we calculate the charge numbers we want?

Jared, I'm just curious, would it be too difficult or add value to show an annual cycle of the Sun/Earth/Moon system?
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Vexman wrote:It seems quoting Miles was more than an appeal to authority. It directly and undoubtedly negates the point about charge binding in Moon-Earth case, while it confirms Earth's field presence beyond magnetopause or 11th ER. Go figure. So for what it concerns me, I'm back to the starting point, scratching my head again. While at it, I'm returning the question back to you: which one is it? Do we have a charge field extending beyond the magnetopause or not? In the case of Venus with 0 magnetosphere, we have it, while in the case of Earth, we don't have it. We can't seem to avoid the results of charge field density calculation, which shows there's actually a point between opposing fields, where we have a charge density equilibrium as we compare them. You can't even imagine how many times I tried to work that over, hoping I could find something to correct it.

As I have been stating all along, it is both. The problem isn't if the earth's charge field can reach the moon, it does, the real problem is what it can do at the moon. You can't treat it as the only field present. It isn't, and it isn't the most powerful, either. However, the only reasons I need it to reach the moon is for exclusion, so I only need the electrical component for that, and then the magnetic component to create axial rotation. Both can be small forces, but perpetually present. But now I have to question that need too, if Miles is suggesting that gravity is not the reason for orbits.

Vexman wrote:On the issue of charge density in relation to the size of emitting surface, I have to disagree. But it's only for the sake of the argument, because I no longer think this issue is relevant to charge binding per se.

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What it means is in my opinion this: one part of Sun's surface equals to 7 parts of Earth's surface in terms of Earth's charge density at its surface. To properly interpret, we need units to measure the surface charge density, i.e. number of photons / cm2, or similar. If Earth has a value of 1, Sun has 7. This may be completely irrelevant, unless one would want to draw a diagram with correct representation of radial lines. Or calculate the charge field strength relative to the arc of emitting body, which may be related to finding the individual charge field limits. But even that requires proper definition of field presence, measured in so far non-attributed units.

I have added the bold to show what I am referring to.

Yes, that is exactly what we need, and that is exactly what we have. While we don't have the number of photons, we have their mass. We express that mass per unit volume, which gives us a density. Just look at the units.

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

The charge densities are expressed in kg per cubic km. So in 1 cubic km, we will find 2.804e+42kg of charge at the sun's surface. It isn't the complete surface, it is only 1km^3 of it. That is why you have to multiply by the sun's surface area to get a total amount of charge emitted, although it probably isn't quite that simple. But that is moot, because we don't need the total emission. We need a unit volume and it already has it in the density value.

If you imagine the surface of the sun, and every square km we place a cube, with each side being 1km, and those cubes poke out into space from that surface, then you have the total emission as the sum of those cubes. But we are only looking at a single cube, not all of them. And each cube contains 2.804e+42kg of charge.

We then let that unit volume move out into space. As it does so, it expands into more space, because it was emitted by a sphere, such that the same amount of charge takes up a larger volume. But we don't want to know the volume, we want to know the amount of charge left in our original unit volume. We want to keep the volume constant and let the density vary. That way, we can say that in 1 unit volume, km^3 in this case, we start with X amount of charge, and after it has moved Y distance, we now have Z amount of charge in that same volume. We do that for both the sun and earth and find the point where they are equal. That is the charge-pause. We are comparing 1km^3 of charge from the sun with 1km^3 of charge from the earth, so we are comparing equal units because we want to look at density.

What you are trying to do with the surface area is handled by the diagram itself. At their surfaces, the sun's lines will be much closer together than the earth's lines and that represents the relative difference in charge density. The lines only represent the direction and nothing else. When the angles are set appropriately, then the lines can be used to determine relative field strength or density of the 2 fields. Essentially, they are contour lines. The closer together they are, the denser the field. You don't need to bring the surface into it because the diagram already does that. You are drawing a sphere, so the curvature is already expressed, and that curvature only affects the drop-off rate of the charge. That is, as the lines move away from the source, the distance between them doesn't grow as fast as it would for a smaller body with more curvature. I think you are trying to get at the curvature difference, but it is already handled in this problem.

In the math, Miles uses the drop-off rate in the 1/d^4 relationship. If you want to make changes based on surface curvature, that is where it needs to be made. The link I supplied earlier was my little trip down that path. Maybe you can see something that I didn't.

LongtimeAirman wrote:It's never been explained? Sorry to be so repetitive, I believe that between the Sun, Earth and Moon, all links coexist at all times. I added the moon and a plot to show the data. The moon’s angular width, seen from the sun, is (of course) 3.669 times narrower than the width of the earth. The sun delivers somewhat less direct emissions to the moon. The earth/moon link is too strong to be overpowered by the sun. The moon is also spinning with respect to the sun; but not with respect to the earth. I believe the earth/stationary moon link is inherently stronger then the sun/ spinning moon link.

If, and only if, the sun and earth emitted the same amount of charge, and were the same size, would that be correct. You are completely ignoring the density of the fields you are using and you are ignoring the curvature of the bodies emitting those fields. You can't use the angular width to determine field strengths like this. Fields are much more complicated than that. As I just explained for Vexman, you have to choose what it is you want to look at in the problem, and keep everything else constant. We want to look at density, so we need to keep the volume constant.

You are looking at the angular width of the moon on the sun, and the angular width of the earth on the sun, and then expecting that to mean something for the moon. It doesn't. All that tells you is that the earth will receive a larger total amount of charge from the sun than the moon will. It tells you nothing about the difference between the sun's charge and the earth's charge at the moon. Essentially, all that tells you is that the earth is larger than the moon. We already knew that and no-one is disputing it.

You say the sun, earth and moon are all linked at all times. What does that mean? Define linked, because for the problem we are dealing with, it is incorrect. I can say that the earth is receiving the sun's charge and the moon is receiving the sun's charge and they represent links. But that doesn't tell us anything about the difference between the sun and earth at the moon. It just tells us that they both receive charge. We already knew that and no-one is disputing it. At least, I am not. It seems like you guys are, but with no reason to do so.

You say the sun delivers somewhat less direct emissions to the moon. What does that mean? Define direct, because all charge is direct. It moves in straight lines. It only reaches the moon or earth if it didn't collide on the way. If it did collide and still reached the earth or moon, then it wasn't going that way to begin with and the collision made it move that way.

You say the earth/moon link is too strong to be overpowered by the sun. How is it overpowered? What makes the earth-moon link so strong?
 

LongtimeAirman wrote:The field lines show that there is always a head-to-head natural direct emission pathway between two bodies – the strength of which is a function involving the addition of both the earth and moon. I’m under the impression that the direct channel between the two object has charge field properties that reinforce or somehow multiply the earth/moon charge link. In other words, the earth/moon link is always passing a two-way charge current greater than either the earth or solar charge fields (at the earth’s orbital radius) alone. The sun is too weak to overpower the direct earth/moon link. However, both the moon and earth also have direct links to the sun. I plotted the earth/moon links in the 12, 1, 2, and 3 o’clock positions of the solar orbit but don't see any value in adding it, the sun's radials at the earth/moon are essentially parallel lines back to the center of the sun.

That's an hypothesis, but not yet an explanation. If we allow the earth-moon link to somehow have these special properties, then why not the sun-moon link? You may think that the moon's field doesn't reach back to the sun, but it doesn't have to. It only has to work within the moon's charge field.

LongtimeAirman wrote:Where did the 7 channels for every one of earth’s come from? If the sun is emitting 7 channels for every one of earth’s, the great volume the sun is emitting into is more than seven times larger at the earth/moon to sun orbital distance. Anyway, curvature is extremely important, and necessary in properly identifying relative field strengths. I agree with Vexman, we don’t have a proper means to compare our fields. Without the proper field density strengths, the radials in these diagrams just indicate direction and the inverse square law.

But it isn't emitting 7 channels to 1 of the earths, it is emitting 84867 of them. Since that is a relative number, we can take it down to the photon level. For every single photon the earth emits, the sun will emit 84867 photons.

We do have the proper means to compare the fields: density. We use a unit volume to make them comparable and we let the density diminish with distance from the source. At any given location, the field with more density is the stronger field, ignoring differing particle sizes. To relate it to the diagrams, if they represented density correctly, then the field with more lines across a given area is the stronger field. Or you can do it with the distance between lines from the same field and compare them to the same in the other field. The smaller distance is the stronger field.

LongtimeAirman wrote:Please feel free to make requests or suggestions. How can we calculate the charge numbers we want?

I already have. I laid out the problems with the diagrams in a post above. I went through how to correct them to reflect the problem at hand. I even gave you the numbers. Currently, those lines tell you direction only. They won't even give you the inverse square law, because that law is about density. Correctly represent density in the diagrams and it will tell you more than it currently is. With the correct diagram, I don't even need to see the earth or sun to know which one is bigger than the other. The lines will tell me that. Right now, they are just arbitrary radial lines. They tell us the field is radial and that is it.

I would like to point out that your original diagram was made to show the angular widths, and that was fine. As soon as it was used to explain charge differences, it was incorrect for those purposes. It is the subsequent usage that I object to and explain how to fix it for that purpose.

Jared Magneson wrote:

Nevyn wrote:Now, I believe that the nuclei must align in order for the upwards charge to be channeled instead of causing a collision. Apparently the downwards charge doesn't get to channel and always collides (or a majority of the time).

How can any charge be "channeled" without a collision?

I had the same thoughts while reading the papers. But we can imagine that some charge does collide, but a lot of it will just go straight through the hole. Imagine we have a wall. That wall has a small hole it it. We have a fireman's hose and spray water at the wall where the hole is. A lot of the water will just hit the wall, but the water that is going towards the hole will not. The edge molecules might collide, but the central ones will not, they will just go straight through.

You do have a good point that charge is not only emitted straight up, and there is no way for those photons to be channeled without collision (which negates the use here because we need to reduce collisions to reduce the effect of the upward forces).

Vexman, I might have led you astray a bit.

Nevyn wrote: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.

That isn't right. I explained that badly. Miles is not calculating the total charge output of the sun. He is calculating the charge density. He is using the total mass of the sun, but a density is always over a unit volume. The bolded part is incorrect, or it is correct, but it doesn't quite say what it looks like. You see, the term total charge density doesn't really mean anything more than charge density, in this context. Precisely because a density is always a quantity within a volume. A density can not represent the total charge output, but it can be used to calculate it. You just have to multiply out the volume in some way that makes sense for the problem.

This is the crux of your problem with the charge field calculations. When in doubt, always look at the units of your numbers. They tell you what your numbers represent and you just have to understand how that fits into your current problem. If the units don't make sense, then the numbers are suspect. I don't care how famous the equation used to generate them. If a dimensional analysis doesn't make sense, then the equation is in the wrong form, at best, incomplete usually, and just plain wrong at worst.

Nevyn wrote:But it isn't emitting 7 channels to 1 of the earths, it is emitting 84867 of them. Since that is a relative number, we can take it down to the photon level. For every single photon the earth emits, the sun will emit 84867 photons.

I would think this ratio shouldn't surprise anyone; the sun is vastly brighter than the sun, in any spectrum. In every spectrum. One can look at the Earth without going blind - we all do it every day. Not so, the sun. And our eyes would be the smallest arc/cross-section of all, relative to the sun's radius.

The moon is bright but nowhere near as bright as the sun. Nothing locally is. But our eyes receive the same arc/cross-section from both, given their balance in charge and gravity (same apparent size from Earth's surface, much like Mercury/Earth/Saturn from the sun's surface). The moon does not blind us. Nor does the Earth, when viewed from the moon.

For as much as I can see, we have different opinions exactly because of undefined units. On the other hand, it's a matter of precision.

Nevyn wrote:"The charge densities are expressed in kg per cubic km. So in 1 cubic km, we will find 2.804e+42kg of charge at the sun's surface. It isn't the complete surface, it is only 1km^3 of it."

Your first sentence says charge densities are expressed in kg per cubic km. They really are within your results, but that is not the correct unit expression, while Miles completely leaves them out. Units come from C=MxD equation. Mass is expressed in kg (or tons), while the density is expressed in kg (or tons) over volume, km^3 in this case - multiplying them gives :

Kg x kg/km^3 -> the result is expressed as kg^2/km^3.

How do we interpret that? This measurement should have its unit, but I'll return to this below. In the end, these units don't play any role in relative comparison, as they get cancelled out if we compare in relative terms, as for instance: body A has x-times charge density of body B.  

The second sentence in your statement implies you can substitute unit of body's volume with its surface. The surface of a cube is not equal to its volume, not in terms of units nor concept. Surface is volume's limit.

Here is the set of data:

.   .   .  .  Earth         Sun
mass        1          333000 Earth's
density     1           0.255 Earth's
surface     1           12000 Earth's

To make it clearer, let's stretch out overall surface area as per equation for sphere's surface:

Earth: Se= 4 π r^2 = 4 π 6371^2=      .       .   Sun: Ss=4 π r^2= 4 π 695510^2=
Se= 510.064.471,9098 km^2         .        .       Ss=  6.078.940.056.315,8994 km^2

Translating that to relative terms as ratio :

Ss/Se= cca 12.000  

We calculated the ratio between the surface sizes, so units get cancelled out again , of course.

Now, let's distribute the relative charge density over the relative surface.

C(Earth) / Se = 1 / 1 = 1     .     .        C(Sun) / Ss = 84900 / 12000= cca 7

The result means that we are looking at the relative charge density over relative surface. If Earth has the value of 1, Sun has 7.

But what are the units of such distribution in absolute measurement? We have charge density expressed in kg^2/km^3 and divide it with surface expressed in km^2 ->  the result would be written as kg^2 / km^5 (or eventually, kg / square root (km^5)). For this to have any sense, we' first need a unit for kg^2/km^3 (charge density), for instance 1 Mathis unit So, when looking at the charge density over surface, we could say it has the strength of 1 Mathis / km^2 (or any other surface unit accordingly).

Where does that come into play? In my opinion, looking at Earth-Sun and their opoosing fields creating a balance point -> the size of Sun's surface which is responsible for earth's charge-pause, is much, much smaller in relative terms than that of Earth's. In terms of percents, it's Sun's little fraction expressing its force over Earth, in terms of both surface and overall charge potential (i.e. 84900x that of the Earth). It's astonishing that such a small part of Sun matches Earth's field at distance.

In his attempt at calculating the distance of charge-pause, Miles inflated Sun's volume to 214 Sun's radii, where it matches inflated Earth's cca at its 11th radii. And this is where I have my issue, since nobody can deny such point of balance.

Imagine Earth shooting photons at the Sun and vice versa. All photons that are hitting the Earth are coming from a small speck on Sun. While Sun's sphere curvature is much smaller than Earth's, its charge density over that small surface is bigger than Earth's. But still, does all that balance out? Is Sun's charge field density that strong, to overpower Earth's in relation to i.e. the Moon?

The mechanism Miles suggests is equally hard to mechanically explain:

"So the photons coming out of the Moon don't feel much spin boost from the Earth's photons (except those going through the Moon). 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."

So somehow Earth's field is dense enough to block Sun's dominant field outside 11th ER radius. Like . . . how exactly?

And lets not forget the fact, that all that mass is roughly 95% of charge going in or out, while there is only 5% of "static" body mass, which could actually account for any kind of material density being visible. In a sense, most of any body's mass is "borrowed" from the field. So the charge field is always there, what matters is field's characteristic - its dominant direction and spin, its coherency. How far can any mass/body reach in terms of dominating the field? What sets the limits? How do we calculate that?

Hmm, I don't think that's right though. Narrowing the charge field down as a simple, perpendicular, UPWARD emission isn't correct. Why? Because the entire DISC of the sun is actually bombarding the Earth. Or the moon. Charge doesn't only move perpendicular straight out, it moves in every direction as it's emitted. The overall PRESSURE or vector is straight out, but the sun appears as a disc in the sky precisely because we're seeing photons emitted at every angle out.

There will be some falloff, sure, meaning the center of the sun from Earth's POV will be where the most photons are coming from and the further out towards the rim, the least (and most oblique angles), but that gradient appears very shallow indeed. In fact, it's almost undetectable.

Look at photos of the sun, even today:
https://sdo.gsfc.nasa.gov/

The site covers many spectrum as well, not just visible. Not only do we not see a gradient, we also see the corona in many of the slides, and solar flares abound. They all contribute to that gradient or rather they all even it out. So the Earth should be feeling charge from the entire disc, not just a tiny section.

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This latest iteration of my so-called earth/moon channel diagram shows an effort at properly balancing the field lines. The space between adjacent source radials at a distance from an object indicates the object’s relative field strength. Closer spaced radials at a given distance from the object indicate a stronger charge field at that location. 1. Solar charge field yellow radial lines are shown. The sun is 150*10^6Km in the 6 o’clock direction, so the yellow radials are close to parallel, fairly constant field strength. Per this diagram, there are 360deg/(4.883*10-3deg) or about 73 thousand solar radials about the sun. The black circle about the blue earth marks the 11 Er magnetosphere radius. The number of radial lines displayed for both the earth and sun were selected such that the distance between earth and the sun’s radial lines are equal at the magnetosphere boundary. That is true when the earth in this diagram is given 36 radials (green). Note that the distance between the two green radials and 2 yellow radials are the same at about the six oclock position of the earth’s magnetosphere circle. 3. The black circle about the moon marks an approximate the moon charge-pause at 9 Mr (Mr = 1.737Km), about 16Km.

Unfortunately, the moon’s radial emission separation is a guess. I looked for some moon charge numbers and found The Moon's Ionosphere as proof of charge recycling, http://milesmathis.com/moonion.pdf.

Miles wrote. I can even explain why the lunar ionosphere peaks at 5km and starts to drop after that. Is is because the ionosphere follows the photonosphere, like everywhere else. In other words, the Moon's charge field remains the major player up to 5km, but above that, the lunar charge field begins to dissipate due to distance from the surface. The density of the lunar charge field falls continuously as we move up, until it begins to be overwhelmed by the charge fields of Earth and Sun. The ions start to get blown away by one or the other.

So I showed the moon with just 8 radials (magenta). For eight radials, the moons ionosphere would equal the sun’s field strength at 16Km, or about 9 Mr (Mr = 1.737Km); the circle drawn about the moon shows that approximate moon charge-pause.

The diagram shows that the Sun’s charge field clearly dominates the space between the earth and moon, but the system is in an orbital balance.

I also believe that the earth and moon receives some charge that has been redirected from elsewhere on the sun's disc. Not all solar photons received come from the small solar angular sliver. Some of the charge from the sliver has also been redirected elsewhere by photon cross traffic. Granted, that additional solar charge does complicate the actual charge calculation; however, I believe that on average, specifying direct solar charge, and ignoring redirected solar charge seems correct to me. Just my opinion.  
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I know that we do have spherical falloff and yes, that matters, but I think you're underestimating the strength of those oblique charge photons. They can only add in, they cannot subtract. It's not a negative-energy thing, which is of course silly.

Anyway, I made an overlay of today's SDO pics, the first and then the second to the last (0193 and Intensity). You can see how the intensity falloff compares to the visible in several way, including the "Subtract" where we see how much charge being discounted if we don't take the entire disc into account, and THEN some.

https://imgur.com/C1VY7aa

The equation is incomplete and is too simple, but still useful if used in the right way. But let's tackle the dimensions. In a quick dimensional analysis, there seems to be too many mass variables. We seem to end up with kg^2. How could that be? It can't be right if the calculated value wants kg/km^3 but the inputs give us kg^2/km^3. But that completely ignores what those dimensions represent, and just treats them as variables. It just blindly assumes that a kg times by a kg must give you kg^2.

So I ask you, and anyone else who feels up to it, to explain what a square kilogram is. Why does a kg * kg give you kg^2? It is easy to see what a square km is. That concept makes sense, but only if you assume that the 2 distances are orthogonal to each other. Each distance must be in a different dimension to the other. But we can't do that with mass, can we? How can one mass be in a different dimension to another mass? It doesn't make sense. Miles discusses this kind of dimensional analysis in one of his papers, but I don't know which one. I think it is about higher order accelerations or something like that.

So, on the one hand, you are right that this equation has issues and is incomplete. However, your attempt to use the ratio of surface areas doesn't make any sense for the problem.

Vexman wrote:The second sentence in your statement implies you can substitute unit of body's volume with its surface. The surface of a cube is not equal to its volume, not in terms of units nor concept. Surface is volume's limit.

No, that is not what I was saying at all, and a little bit further on I explained exactly what I was doing:

Nevyn wrote:If you imagine the surface of the sun, and every square km we place a cube, with each side being 1km, and those cubes poke out into space from that surface, then you have the total emission as the sum of those cubes. But we are only looking at a single cube, not all of them. And each cube contains 2.804e+42kg of charge.

I am not substituting surface area and volume, I am using the surface to place the volumes. We have a density value and I am showing how it applies to the problem. The part sticking out from the surface represents the time differential. It is the dt since mass of charge is not really a kg, it is a kg/s or some other unit time. I am representing the problem because we have a whole bodies worth of emission expressed at the surface.

Surface area is expressed as a square distance. It is 2-dimensional. A volume is expressed as a cubed distance. It is 3-dimensional. A 3D entity is a 2D entity extended into the 3rd dimension, at least when discussing squares and cubes. I am using that relationship to say that each square km of the surface can have a cubed km of volume associated with it, or attached to it. We take 1 side of the cube and make it equal to 1 square km on the surface, and let the rest extend out into space.

You are assuming that the value 84867 represents the total output of the sun, but it does not. It only represents the charge density of the sun in terms of the earth's charge density. You may not realise that you are making that assumption, but your math does. The moment you divide by the surface area, you are assuming that you have a total output. You are assuming that the number being divided represents the complete surface, otherwise there is no point to dividing by that surface. Dividing by a surface area is a way to get a 2D density in the same way dividing by a volume gets a 3D density. So it is wrong to think that you can divide a density by a surface area to get a density. You divide a quantity by a surface area/volume to get the quantity per unit area/volume.

I believe you are trying to find the amount of charge emitted at the earth. The small section of the sun that points to the earth, but your math isn't doing that, and we don't need to anyway. The density is already in the right form to do this. It represents a unit volume, and we use the same volume for the earth so that they are comparable. We say that the sun emits X amount of charge in Y volume of space. The earth emits Z amount of charge into Y volume of space. Since the volume is the same, we can compare the strengths of those fields. We then just adjust for distance travelled where the density will vary but the volume will remain constant.

What you should be trying to do, is find the surface area of the sun that points at the earth. Then multiply that number by the charge density of the sun at its surface to get a total amount of charge (per unit time) over that section of the surface. Because we are starting with a cone that extends out to envelope the earth, we can't reduce the amount of charge. The volume is growing with distance from the sun, but the amount of charge is not. We are keeping the amount of charge constant and letting the volume vary. By volume, I really mean the distance between the lines from the center of the sun to the edges of the earth. I don't know what to call that. We are looking at a volume, but that line shows the variance more easily.

Now we have to look at it from the earth's perspective. How can we determine the cone of space that we want to keep track of? We want to know the point where the 2 fields are equal. The sun's field is represented by a cone that is getting larger the closer it gets to the earth. We need to know how much of the earth's charge interacts with that volume. But the farther we move away from the earth, it is a shrinking volume of space, so the amount of the earth's charge enveloped by that volume is changing. The earth needs to vary both the charge and volume. A very complicated problem and prone to error, but it can be done.

Basically, if you already know where the balance point is, then you can work out how much of the earth's charge gets there, because you know the volume of space that the earth is allowed to emit into for this problem. Miles' method of calculating the charge-pause has the same problem. He can calculate where it is, knowing where it is, but he is trying to show why it is more than where. To calculate where it is with no other knowledge is more difficult, and can only be done by exhaustion (basically try some distance, if not equal, try another). I had a go at calculating the charge-pause directly, some time ago, and should probably try again. I'm sure there is a way to do it, but I didn't find it the first time. It is easy to run yourself in circles with this stuff.

Vexman wrote:The mechanism Miles suggests is equally hard to mechanically explain:

"So the photons coming out of the Moon don't feel much spin boost from the Earth's photons (except those going through the Moon). 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."

So somehow Earth's field is dense enough to block Sun's dominant field outside 11th ER radius. Like . . . how exactly?

Miles is being very careful with his language in that quote, and you need to be aware of it to understand him. Note that he says a sort of wall. He doesn't say a wall. Then he says keeping the Sun's photons from defining the ambient field. The bit that is easy to miss is defining. He is not saying that either field has stopped or that they have no charge presence or even that they have no strength. He is just saying that they are not the strongest field at this location. The ambient field is location dependent and is defined by the strongest field present. So if you are 10,000km above the earth, then you are in the Earth's field, it sets what we mean by ambient field. The sun's field is still present. The moon's field is still present. Even all of the other planets have some presence. But all of those other fields must be used after the earth's field in any problem at that location.

I am no help with the first sentence, sorry.

Airman, that diagram is heaps better. It gives so much information now. It would be good to get another version with all charge fields in the same line. If the earth and moon were in-line with the sun's field, we could differentiate them better. Basically turn the yellow lines 90°. That gives us both orientations with respect to the sun. Putting the moon behind the earth, from the sun, would be another good one, but not really necessary for this discussion.

With respect to the moon's charge strength, have a look at The Moon Gives Up A Secret. That should have some data for the moon.

Airman, I think that the lines for the moon should be much farther apart. Start at the moon, follow a straight line to the earth, but stop at the circle that represents the moon's charge-pause. The tangent to that point on the moon-pause, extended up and down, then crosses the green earth lines. From both of those points, draw a line back to the center of the moon. That is the angle for the moon's charge field, because that is the angle where the earth and moon fields are balanced. The angle will be close to 3x your current angle for the moon. You already have magenta lines that almost cross at the right points.

Sorry Airman, I've just realised that you made the moon's charge-pause equal to that of the sun. I was thinking the earth. My mistake. Carry on, nothing to see here.

Want to say again that I don't negate or argue comparison of charge density over unit volume.

I stared at those units for days, Nevyn, and couldn't make any sense out of it.  There is no error in sticking blindly to the fact that kg multiplied by kg produces the result of kg^2. I didn't invent that, it's a rule of mathematics and not an assumption. There's no arbitrary way to decide which units you need or want, equations and transforms require that you stick to the rules not only in regard to numbers, but units too. You seem to be annoyed by me bringing this up, which was not my intention. I do stand behind everything I wrote about in my previous post, including the units though.

I'm not sure that I know what kg^2 stands for and how to properly interpret it, which I admitted at the beginning of my previous post. Maybe Miles should answer to your question as he postulated the formula. I'll give it a try with this post.

Quoting Miles again, my emphasis:

"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. So we find by simple math that the charge field density is vastly different than the material density. We are finding the charge field density at the surface of the Sun, so we must sum all the mass “behind” that surface. All that mass emits charge photons. "

So, as you can see - he too wanted to seek a charge density at the surface of the Sun. Miles'  idea how to do it is right there - by summing all the mass "behind" that surface.

Well, it's true half-way in, since the surface is just an envelope around the volume, it has no volume by itself, which is why it's expressed in 2 dimensions. If you follow exact Miles' suggestion, it implies that by summing the mass within body's boundary (i.e. below or behind a surface as in volume), you get the result of that summation at the surface. In his calculation, he never looked at or calculated the surface area. So when did the distribution of volume over its surface happen? In my opinion, it didn't happen. His idea was different and it was about inflating the volume to certain radius and calculating the charge density drop-off according to 1/r^4. All charge behind the surface is what he was looking at - the sphere's volume with its charge density at inflated radius. But that doesn't say anything about the charge density at the surface, which is in Sun's case 12000 times bigger than Earth's surface.

Anyway, if he allowed for the substitution of charge with mass at the beginning, we can do the same afterwards. In this light, final units kg^2/km^3 can be read as : one kg of charge per each unit of body's material density (kg/km^3). It took me a whole week to write down that previous sentence. Looking at the units after the distribution over the surface, kg^2/km^5, we can possibly interpret it as : one kg of charge per each unit of body's material density, distributed over one square km of body's surface.
I don't know if that makes any sense, but it's my attempt at explaining it. So far, it seems as I'm the only one who tried to do so.

I'm not annoyed with you for bringing that up. Maybe a little annoyed at myself for not following through and seeing that myself. As I've said before, I think there are problems with C=MD, it is too simple. Maybe we can work out a better equation to use. I don't know.

Vexman wrote:I stared at those units for days, Nevyn, and couldn't make any sense out of it.  There is no error in sticking blindly to the fact that kg multiplied by kg produces the result of kg^2. I didn't invent that, it's a rule of mathematics and not an assumption. There's no arbitrary way to decide which units you need or want, equations and transforms require that you stick to the rules not only in regard to numbers, but units too. You seem to be annoyed by me bringing this up, which was not my intention. I do stand behind everything I wrote about in my previous post, including the units though.

But it is an error if the assumption is wrong. And it absolutely is an assumption. It doesn't matter if it is a rule of mathematics. If the concept doesn't make sense, then at best it is an assumption. This is what happens when you let mathematicians do everything. They run away from reality the first chance they get and hide in the math. Physicists don't have that luxury. They are absolutely bound by reality. At least, they used to be. But, yes, I will be ridiculed for saying that and you will be thought correct for following dogma. I don't mind. My statements make sense, the accepted position does not. I'll take sense over authority any day.

Vexman wrote:I'm not sure that I know what kg^2 stands for and how to properly interpret it, which I admitted at the beginning of my previous post. Maybe Miles should answer to your question as he postulated the formula. I'll give it a try with this post.

I guess my question about that looked like I was throwing it back in your face, as they say, but I was just asking the question to help you see that it doesn't make sense. I wanted you to think about it and try to find the physics behind it, not the mathematics. If the math doesn't reflect the problem, then it is useless. Reality comes first, the concepts come second, and the math comes a distant third.

I might be able to clear this up by showing that the equation uses the sun's mass, but really wants the mass of the charge emitted by that mass. If mass is really kg/s, say, then a mass of a mass might be a kg/s^2. It is a second measurement of something that is already a measurement over time, or a second order measurement. That would imply that it is a derivative. We are bringing calculus into it. It might be an avenue worth pursuing. I don't know. I'm just trying to see it from different perspectives to gather some light on the problem.

Vexman wrote:Quoting Miles again, my emphasis:

"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. So we find by simple math that the charge field density is vastly different than the material density. We are finding the charge field density at the surface of the Sun, so we must sum all the mass “behind” that surface. All that mass emits charge photons. "

So, as you can see - he too wanted to seek a charge density at the surface of the Sun. Miles'  idea how to do it is right there - by summing all the mass "behind" that surface.

Yes, but you have to interpret that statement correctly. He is seeking a charge density, so he is seeking a quantity spread over a unit volume. He is not seeking a quantity spread over the surface. The quantity is at the surface, but it is not covering that surface. You are implying that Miles wants it to be spread over the complete surface, but the word density says otherwise. We can argue whether the equation used to get that density is correct or not, but once we have a density, then we can't say that it is something else.

Vexman wrote:Well, it's true half-way in, since the surface is just an envelope around the volume, it has no volume by itself, which is why it's expressed in 2 dimensions. If you follow exact Miles' suggestion, it implies that by summing the mass within body's boundary (i.e. below or behind a surface as in volume), you get the result of that summation at the surface. In his calculation, he never looked at or calculated the surface area. So when did the distribution of volume over its surface happen? In my opinion, it didn't happen. His idea was different and it was about inflating the volume to certain radius and calculating the charge density drop-off according to 1/r^4. All charge behind the surface is what he was looking at - the sphere's volume with its charge density at inflated radius. But that doesn't say anything about the charge density at the surface, which is in Sun's case 12000 times bigger than Earth's surface.

There is no distribution of volume over the surface because that is not what Miles is doing. He is taking a unit-volume located at the surface and then looking at what could emit charge into that volume. It isn't only the mass at the surface that emits into it, it is the whole mass of the sun. Right now you are on the surface of the earth. In no way does that mean that you cover the complete surface of it. It just means that you are located somewhere on that surface. Now look around you, is there any place within the earth that could not emit charge at you? You can take a single proton on the opposite side of the earth from you, and we can still assume that some of its charge could reach you. That is what Miles is getting at. The complete mass emits and some of that will reach our unit volume. The equation to calculate the density doesn't get into that, and it can probably be ignored for this problem, being treated relatively rather than absolutely, but it is something that needs to be dealt with if we want to find the absolute amount of charge.

Vexman wrote:Anyway, if he allowed for the substitution of charge with mass at the beginning, we can do the same afterwards. In this light, final units kg^2/km^3 can be read as : one kg of charge per each unit of body's material density (kg/km^3). It took me a whole week to write down that previous sentence. Looking at the units after the distribution over the surface, kg^2/km^5, we can possibly interpret it as : one kg of charge per each unit of body's material density, distributed over one square km of body's surface.
I don't know if that makes any sense, but it's my attempt at explaining it. So far, it seems as I'm the only one who tried to do so.

I think the better approach is to fix the density equation. We know Miles wanted a density, but the equation either doesn't give us one, or it has dimensional issues if we follow current assumptions. This problem can be broken down into 2 parts (at least): find the density; and use that density. Either part can be changed without changing the other, but we can't change that it is a density.

On a personal note, if I thought you were on the completely wrong track and not understanding a lot of the problem, then I probably wouldn't even post about it. I might, if I thought I could help you, but I wouldn't spend a heap of time on it. I wouldn't waste my time when Miles papers are there to be read. I don't want anyone to think that I see it as beneath me, but there have been posts made to this site that I have thought were so far off the track that it would take too much effort to fix. If I have better things to spend my time on, then that is what I will do. I don't need to fix everyone's misunderstandings, but sometimes I do make an attempt. But I see potential in you. I've enjoyed you questioning me. I look forward to the responses. I like the way you are thinking about the problems. I like a challenge and am always looking for a better understanding. I know that I can be wrong. I know that I can miss things. I know that I can forget about things. I don't have all the answers and sometimes I am trying to justify things, such as the dimensions above. At the same time, though, I can usually see what are the key points and what is flexible in a given problem. In this case, the density is set in stone, but the equation to get it is not. I don't mean the density as a value, I mean it as a concept. It is the key part of the algorithm.

So, please, try to take what I say as genuine inquiry and not intended to be malicious. Even when I am heated, or frustrated more likely, about something, I try to remove it from my posts. I try to remain focused on the problem at hand and not get personal. Maybe I don't always succeed. Maybe I don't use the right words. Maybe I am saying things in a way that make sense to me, as an Australian or even as a software developer, but don't to others with very different backgrounds. We do have an international community here, and that can make it difficult to convey information sometimes. I might see something as comical or light-hearted, but others take it as an attack. I often re-read what I have written after a bit of time has passed to see if I am making sense or not. Sometimes even I can't see why I thought a particular thing. It might come back to me eventually, but it means I have not written it in such a way that it conveys the complete picture, or I might see how it could be misunderstood and I haven't been clear enough. That is on me. The onus is always on the writer, never the reader, to express things in a way that they can be understood. However, the reader does need to make sure that they are hearing what the writer is saying. If I use the term density, then I expect the reader to know what that is and how it can be used appropriately. If they don't, then they need to find the definition and re-think the situation. It is easy to get confused with all of this. I don't hold that against anyone, because I might be one statement away from seeing my own confusion.

I had a wall of text already written in reply, but I don't think it would do any good to this discussion.

Just two condensed points: there is absolutely no error in applying rules of multiplication to both the numbers and their units. To me, it matters a lot. It's the reason I like math, there is no partial agreeing with it, it's all or nothing. By the way, I'm not a mathematician by profession nor did I study it. And I'm no physicist either. But I know that rules of multiplication are not a dogma, they're all about logic and precision. So let's rather argue about the concepts, using arguments.  

Material density is defined as mass over unit volume, Density (sigma) = m / V.  

In Mathisian physics, Miles has proposed charge density as C= m x D , which can be written as C=m x m/V = m^2/V . So is this a conceptual error or not?

If not, can it be interpreted like: unit of charge per unit of material density?

If it is an error in concept, how is it possible we were able to calculate the distance of Earth's charge-pause using this concept? Coincidence in set of numbers ? Is there any other case where we can verify it?

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kg^2/km^3

What the heck is it? I’ve been barely able to understand the discussion, let alone contribute. Still, you may appreciate this fine source.

Miles wrote:Next you calculate the repulsive force caused by the radiation. This is a real force and it is caused by direct collision. To calculate it, you need to know the density of your field. In other words, you need to know how many of these radiated smaller objects are impacting the second object, and how much they weigh. The amount of radiation is directly determined by the radiating object. The more atoms it contains, the more radiation it will be emitting, since the radiation comes right out of the atoms directly. We already know that. A more massive object will create a more massive field, since the field is a direct extension of the object.

That’s from Why the Sun and Moon have the same Optical Size Proof of my Theory. http://milesmathis.com/third9.html.

If I interpret that correctly - a big if, Kg*Kg may be the mass emitted by the source mass.

The Third Wave is about expansion theory, it takes a bit of effort to translate that to charge binding. The paper has several points worth mentioning. For example,

Miles wrote:The Sun and Moon always look the same size from the Earth and from nowhere else.1 Given that, and given my mechanics, what I said before is strictly true. The Earth is accelerating toward all objects at the same rate.
     If it is not approaching those objects, it must be because the other motions are preventing it. The Earth is not approaching the Moon or the Sun. The average distance is constant. Therefore, the Sun and Moon must be repulsing the Earth in the same amount.

That tells me that the earth feels the same repulsive charge density from either the moon or sun. Based on that fact, here are two more charge channel diagrams with what I believe is a proper Sun/Earth/Moon charge field diagram. The first is with the sun in the center, and the earth/moon at 12 o’clock, and the second image shows the earth/moon at 3 o’clock. The earth, the blue circle is surrounded by the 11Er magnetosphere, where the earth’s emissions match that of the sun. The moon radials, 186 each 1.931 degrees apart are now shown to be the same strength as the sun at the earth.





Agree/disagree?
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If the rules of math are all about logic and precision, then we should be able to see that logic and precision. If you can't make sense of a concept, such as kg*kg=kg^2, then where is the logic? What is the justification for doing that? You say that there is no partial agreeing with it, but it is not dogma. Well, that is the definition of dogma. Your proving my point. Math isn't perfect, and it isn't complete. Mathematicians love to talk about how they are uncovering truths because that makes it appear irrefutable, but it doesn't make it so.

At the end of the day, the math has to reflect the concepts in use, which have to reflect reality as we know it. That overthrows another standard mantra of physics and math: that the math is everything. It is a distant third at best, and any self-respecting physicist worthy of that title would agree. The first thing a mathematician will do is run away into pure math, so they can avoid reality. A physicist uses applied math, so they can't (or shouldn't) hide in pure math at any stage. They do need to make sure that they are applying the math correctly though. That's a quirk of math, it can be correct as math and still be used in the wrong way, giving incorrect results.

I'm no expert in math. I was always good at it, and studied some at university, but didn't go as far as I would now like. I'm no expert in physics either, there is plenty I don't know, but I try to make sense of things as best as I can. I am an expert in algorithms, though, and that allows me to see things a bit more abstractly. It allows me to see what is concrete and what is malleable. I've spent enough time trying to make sense of Miles work that I can usually see how he is putting things together and how they relate to each other. I don't treat the math as a rigid structure, I treat it as a bag of tools where I grab the most appropriate tool for the job. If there isn't one, then I try to find it. If I can't find it, then I use lower level tools to create it. Well, I make an attempt.

In order to work out a new density equation, we need to make sure that it still works with the rest of the algorithm. To ensure that, I will setup a webpage that calculates the second part of this algorithm: the equality of charge density. The actual density equation will be malleable. I might even try to allow the user to specify the density equation in text and it will use that to make the calculations. Basically, I want a test-bed so that I don't have to make so many calculations every time I change the density equation.

When that is done, I will create a new thread here so that we can discuss it without derailing this gravity thread.

I may be looked at as some kind of heretic for saying my most sincere opinion, but I don't care so I'll say it anyway.

Where's the logic in kg^2? There is no logic in kg over two dimensions. Did we do the procedure of arriving at it correctly, is math behind it correct, with no errors? Yes. Then we need another concept of C. I want to be proven wrong before I apologize for having such belief.

Or, I need to accept what Miles has said about the charge density:

"So we find by simple math that the charge field density is vastly different than the material density."

I will take this question of units directly to Miles and ask his permission to share his response here. Hopefully, he'll both reply and allow sharing.

I apologize for derailing this thread. I won't happen again.

Some bad news, which we may be able to spin into something good. Miles has failed to calculate the charge-pause. I'm a bit disappointed. I can see where the mistake is made, but a mistake it is.

I created a page to find the distance between two bodies given their radius, mass and density. Using the known distance between them, the app will find the point along that distance where the fields are equal. It presents the distance from each body and then expresses that distance in terms of the radius of each body.



You can use it here: https://www.nevyns-lab.com/mathis/math/charge-field-equality.php

The default values are for the Sun as Body 1 and the Earth as Body 2. I am using different raw values for radius, mass and density, so we expect some discrepancies in the exact numbers. The image above shows the results of the calculations.

If you take a quick look, you will find the 11.90 value that we expect for the earth, on the right-hand side. On the left we have 204, which we can say is close enough to the 214 that Miles mentions in the paper. It looks like a match, but it isn't. It isn't even close to a match. I could find alternate sources for the properties of the bodies and get closer to Miles actual numbers, but it is pointless. That isn't what the problem is. The problem is that this calculates the Earth's charge-pause as being 11.90 Sun radii from the Earth.

I checked my code and made sure the right values were being put into the right fields, but I'm afraid they are. I checked it 3 more times, and they were still being put into the correct fields.

What does this mean?

It means we have room to find a better charge density equation. After reading Vexman's response this morning, I made an attempt at deriving the density equation. As I did so, I was hoping that there was a way that it would be done without losing the current number relationships. I expected the values to change, but as long as they changed for both bodies in relatively the same ways, it should all work out. As I went along, though, I couldn't see it and I kept thinking that the changes I want to make are going to break everything.

Now we know that we don't have to live up to the current numbers. That gives us the freedom to re-write the charge density equation and test it against the charge-pause.

Vexman wrote:I may be looked at as some kind of heretic for saying my most sincere opinion, but I don't care so I'll say it anyway.

Where's the logic in kg^2? There is no logic in kg over two dimensions. Did we do the procedure of arriving at it correctly, is math behind it correct, with no errors? Yes. Then we need another concept of C. I want to be proven wrong before I apologize for having such belief.

I can only speak for myself, but I won't think that of you, no matter the opinion. I've disagreed with Miles plenty of times, so don't worry about that. I also don't want an apology if you are wrong. I just want you to get a better understanding, or I do if I am wrong. That is my only goal.

Vexman wrote:Or, I need to accept what Miles has said about the charge density:

"So we find by simple math that the charge field density is vastly different than the material density."

I will take this question of units directly to Miles and ask his permission to share his response here. Hopefully, he'll both reply and allow sharing.

You might want to read my last post before you do that.

Vexman wrote:I apologize for derailing this thread. I won't happen again.

My opinion is that this has been a good discussion and I've enjoyed it. While we have gotton slightly off track, they are related to the gravity paper. I don't think it has been derailed so far, but if we get into re-writing the density equation then that would be off-topic.

Well, the email went out before your reply was posted. It won't hurt to know Miles' opinion.

Your previous post was a big a-ha moment for me. The relative size of Earth's radius was not correctly applied in Miles' attempt. Now you've proved it by inserting absolute numbers into the equation. Thank you for doing so.

Thanks for all the discussion here, guys, and let's keep digging at this of course.

@LongtimeAirman: do you think you could put labels on those diagrams? Honestly I'm having a tough time figuring them out. I'll re-read your descriptions again though. Just an idea!

Nevyn wrote:
It isn't only the mass at the surface that emits into it, it is the whole mass of the sun. Right now you are on the surface of the earth. In no way does that mean that you cover the complete surface of it. It just means that you are located somewhere on that surface. Now look around you, is there any place within the earth that could not emit charge at you? You can take a single proton on the opposite side of the earth from you, and we can still assume that some of its charge could reach you. That is what Miles is getting at. The complete mass emits and some of that will reach our unit volume. The equation to calculate the density doesn't get into that, and it can probably be ignored for this problem, being treated relatively rather than absolutely, but it is something that needs to be dealt with if we want to find the absolute amount of charge.

Indeed, I think that's what I was trying to get at earlier with my claim that the entire body needed to be considered. The disc of the sun, the disc of the moon, are just what we SEE, but everything in the entire body needs to be accounted for. Not just perpendicular charge from a given place on the surface.

But all that doesn't really help me make sense of charge-binging gravity, personally. Still awaiting Miles' input or next paper on this one, I'm just as in the dark as I was when we began.

I do appreciate Vexman and Nevyn's frank discussion, and from an outside perspective (knowing both of you pretty well) I don't sense any real ire or conflict - so I hope we can remain how we are, critical but friendly - since I myself value the shit out of both of you. Glad you could join us here, Vexman. This is a really tough topic so far and I appreciate your input, and the conversation all around from everyone. I don't think you derailed the topic any more than I did, or anyone else - we're all trying to figure out the same thing (or falsify it) from different angles.

.

Hi Jared, of course, it's a rough approximation at present. I don't hear any criticism yet. I hope the description is acceptable. It is based on angular widths. Given the two facts, that the pairs of earth/sun fields balance at the magnetosphere and the moon/sun fields balance at the earth, I don't believe the ongoing discussion changes this image - much. It should be easy enough to create an application that would allow one to select specific directions or distances and allow closer views.

P.S. Replaced image.
.

Vexman wrote:Well, the email went out before your reply was posted. It won't hurt to know Miles' opinion.

Your previous post was a big a-ha moment for me. The relative size of Earth's radius was not correctly applied in Miles' attempt. Now you've proved it by inserting absolute numbers into the equation. Thank you for doing so.

I haven't proven that. I have disproved Miles' claim to have calculated the magneto-pause. To prove your claim, we need a density equation that does what you think it should, and it satisfies the charge-field equality algorithm. If you can supply that equation, I can now test it easily. This disproof shows that the equation has problems and is ripe for replacement, but it tells us nothing about what is wrong with it. That doesn't mean you are wrong, just that it isn't proven yet. Also note that the problem can be in the charge-density drop-off relationship too.

I've updated the page to allow you to enter the charge density and density drop-off equations. If you want more variables in your equations then let me know and I will make them available.

The math expression is compiled by a JS library called MathJS. If you want to know about the advanced features, such as functions, then consult their documentation at https://mathjs.org.

It will usually be something like this: sqrt(3^2 + 4^2)

I want to believe Miles would've seen his own error if he applied the relative radius properly. If sun's R was used for the Sun while applying 1/r^4, the same R should have been applied to calculate for the Earth. Or vice versa, R of the Earth should've been used for both bodies. To set the relative ratio in proper relationship was crucial here. Your absolute result is showing it in my opinion. Don't forget he arrived at the balance point by exhaustion, or trial&error method, inserting relative values of radii until he arrived at the proper balance. But still, I think he would've realized his error if correct relative sizes were applied. The right thing to do was taking absolute numbers and checking the relative result by running them in equation, which is what you did.

I've received the answer from Miles and his permission to share. Permission came with a disclaimer, saying he didn't study the subject of charge density units in depth so his interpretation of its units should be looked through that lens. Then I told him about the whole depth of the problem and the error in his EM calculation and my view of C=MxD concept. I also decided not to share his units interpretation until he has a chance to correct or address this issue properly. I want to believe it's fair to let Miles first realize the error he made before jumping at him for his interpretation of C-units. And I also hope nobody here will hold it against me for making such decision.

I would expect we're going to see an update on this in the form of a new science paper.

I'm a bit concerned about the charge density drop-off equation and the way I have used it and if that is how Miles has used it. I am using it like this:

density at distance = density / distance^4

The distance is always an absolute value. That is, if the distance between the bodies is km, then that is what the distance used in the drop-off equation is. It is never scaled to the radii of either body. Now I'm starting to remember why Miles scaled it. I checked the paper, but it doesn't talk about it, maybe I read it in another, maybe I came up with it myself to explain it, I don't know. I think the idea is that the C=MD equation gives a charge density at the surface, or stated another way, at 1 radii. So the drop-off equation may need to use those units instead of absolute units. I always wondered how the curvature was being expressed in the drop-off equation and now I see that this is how.

So I may need to use it like this:

density at distance = density / (distance / radius)^4

Ill see how that goes and get back to you all.

Good news, that solved the problem.



The rest of the numbers match too, so I think I've fixed the problem and I am representing Miles math correctly.

Now we can debate about whether we are allowed to do that.

To check against other planets, I have added the ability to specify the planet used for Body 2 as a URL parameter. Just add ?planet=venus to the URL, and replace venus with the planet you want to use.

https://www.nevyns-lab.com/mathis/math/charge-field-equality.php?planet=venus

I don't understand your postulate here. Distance is expressed in km, eventually. If you set it as a relative term, then distance is expressed in radius. But I think that you should set both distances to be relative to a single radius you have chosen to compare it with. For instance, if you set distance as l/r, the radius being used for Sun should be used for Earth as well. I think that was where Miles went the wrong way.

I'm attaching my excell table, with calculated values in absolute terms. They are equal to numbers in your previous post, where the absolute values show 2-field balance at roughly 203 Sun's radii from the Sun and 11,9 Sun's radii from the Earth:

You have to use the radius of each body for the calculation of its own charge field at a distance because the original value, C=MD, is relative to that radius. C is already a relative value. C represents the charge density at 1 radii. We can't ignore that 1 radii and treat it as any old number. The drop-off equation needs to treat it as 1 radii, so we divide the absolute distance by the radius of the body to make the distance relative to the same radius. We make them of equal rank so that they can be used together.

The reason I missed this drop-off problem is that Miles talks about it as a 1/d^4 relationship. That is misleading, though, because it ignores the need to make d relative to the radius.

The drop-off equation has to express the drop in density, but it needs to do it in a way that represents what is emitting that density. A very large body will drop off at a rate slower than a small body. This is how the curvature of the body is expressed. A smaller curvature means a flatter field which drops off at a slower rate. Airman's diagrams are showing that by the distance between the field lines. The yellow Sun lines don't really change the distance between them, but the Earth's and Moon's do. They spread out quite quickly.

So we can't use a common radius because that would be ignoring the size differences of the bodies.

So, in relation to units, that leaves us with the knowledge that C=MD is not calculating a density exactly, it is a density at a distance. We can't think of it as a strict quantity over a volume. It has that property, but it has more too. It is an intermediate value, rather than some absolute concept. The density drop-off equation then completes the calculation and expresses the curvature of the radius and how that effects the density.

Nevyn, I calculated for both Earth and the Sun their absolute values of C. Masses are expressed in kg, while material density is expressed in kg/km^3. The distance in my calculation is expressed in km, I only represented them in radius length at the end, for easier understanding.

For instance, the value of Sun's charge density at 214 Sun's radius is in my calculation written as : Cs / (214*695699 km)^4.

The ratio between absolute values should be equal to ratio between relative units. But it's not if you do the check against your most recent update.

My recent updates are calculating that like this: Cs / ( 214*695699/695699 )^4, to put it in the same way that your equation is. Basically, you don't need to multiply by the radius because 214 is already relative to the radius.

The way I interpret all of this is that these equations are showing us a small piece of a larger puzzle. The units don't work out, the equations are doing different things to the values and it looks a bit weird. Maybe there's a way to work out the dimensions. I don't know. It looks a bit daunting. I think the problem is that it can't be broken down into isolated equations. The equations have to work with the concepts of the problem. Those concepts are complex and it can be difficult to see what the equations are doing and how they are doing it.

When I say isolated equations, I mean general equations that you can use pretty much anywhere. These equations are not that general. They are written for a specific problem and can only be used in similar problems. They are expressing fields and curvature and distance and density, all at the same time. That makes the numbers very specific to the problem.

I'm posting the table again, only absolute units were used to calculate:

As soon as you calculate Cs and Ce, you are not using absolute numbers anymore. You are trying to force C into an absolute number, but it is not because the equation is not designed to give you an absolute number. Cs is relative to the sun and Ce is relative to the earth, so you can't expect to treat them as absolute numbers by giving them absolute distances. The drop-off equation doesn't want absolute distance, it wants a relative distance because the density it is using is relative. When given a relative distance it can then express the curvature of the sphere that created that density.

It seems like you are struggling to see how we can compare two values that use different radii. They seem to be out of sync with each other and therefore not comparable. However, the drop-off equation is what makes them comparable. It converts them into something by using the radii and whatever that is creating is comparable.

I'm not following you anymore. Radius is an expression of length or distance, and it should be expressed in according units (m or km). But that is not the issue here. Cs is not relative to the sun if you use absolute units to calculate it. Why should it be? It expresses Sun's charge density, which is expressed in kgˇ2/km^3. There is nothing relative to the sun here when using absolute numbers. And there is no requirement within the C=M*D that would say anything about prohibiting the use of absolute numbers to calculate the result.

Two different radii should be fixed in a ratio if we want to compare them in relative terms. Rs/Re is 103,26 , inversely Re is 0,00968 that of the Sun. So when comparing them, we need to choose one and stick to it all the way through relative comparison. We can't be switching radius when comparing them in relative terms.

I'm honestly a bit lost in how this would affect the charge-binding gravity theory, myself. We still have a huge acceleration vector down and a tiny vector up. Lift on a wing still works so we know that it's still a tiny vector up, yes?

Of course it's an interesting physics problem, finding the charge-pause of the Earth properly and dealing with that - not saying I mind side-railing the thread, but so far I can't see how this helps us make any more sense of Miles' newer theory than we had previously. Which to me so far is none, still.

I'm going to attempt to diagram his theory in animation as I see it, and perhaps you folks can show me where I go wrong. Sort of reverse-engineer it, if I can. But it might take awhile.

You're treating this as a purely mathematical problem and ignoring the physics. You expect the equation to give you absolute numbers just because you gave it absolute numbers. You are completely ignoring what the equation is trying to do and just treating it as a function. It isn't a function. It isn't giving you a simple value. It isn't giving you a density, but a density in action, in a way.

These equations are compressed. They are engineer's equations, as Miles would say. They are not in their expanded form, but you are treating them as if they were. You think that they can give you perfect knowledge, but they are hiding some things from you. To figure out why C=MD works, we need to unwind it and find its most expressive form. If we did that, you would see that there is more to it than just the mass and density. Somehow it all cancels down to those values, or at least they give us a value that works. It might not be the exact value, but as long as the relationships hold, then it doesn't matter. By that I mean that if it alters the sun's value in the same way that it alters the earth's value, then the actual numbers don't matter as much as the results that they give. They are relatively the same because whatever is happening is happening to both sides in the same way.

In mathematics, we often do something to an equation to help us derive it. For example, we might multiply both sides of the equation by 2. We can do that because it is doing the same to both sides and the equality still holds. That is what I am trying to get at in the last paragraph.

I'll try to break down the problem to see if it helps us see it more clearly.

We have 2 bodies that are emitting a field of particles. We want to know where those 2 fields are equal. So, in the first instance, we have something like this:

F1 = F2

where F stands for field.

Now we need to expand those F's to fit the problem. To do that, we need to look at how those field are created. We find that they are being created by a sphere. So that needs to be represented somehow because a density created by a sphere will drop-off with a 1/l^2 relationship (l = length). We are actually talking about very large bodies, though, which brings Einstein's Relativity into it a bit because it takes time for the charge to move to the other body, or the point of equality. That brings in another 1/l^2 relationship and we get a 1/l^4. That relationship assumes a radius of 1. That is probably the crux of your current problem right there. The relationship assumes a radius of 1, so to use that relationship, we have to accommodate that.

To use that relationship we have to give it lengths that mean something to it. An absolute length won't cut it. It doesn't give the relationship what it needs because it is the relationship that creates the need to be relative to the radius. The relationship not only has to represent the drop in density, it also has to represent the size of that which is creating the density. That size matters, a lot.

However, we need to use an absolute distance, so the drop-off equation has to take care of that and make sure that it converts it into a form that the relationship wants. To do that, it shrinks the distance down to a size that represents a unit sphere. Now we can use the 1/l^4 relationship that absolutely wants to work with a unit sphere. So the equation becomes 1/(l/r)^4.

dropoff = d/(l/r)^4

So we currently have:

F1/(l/R1)^4 = F2/(l/R2)^4

Now we have to expand out the F terms, which is tricky. What do we know about them? We know that it represents some kind of density. We know that it represents the emission of the body, not the body itself. We know that whatever it is, it must be located at the surface of that body. We know that because the density drop-off equation wants it to be there. That is the relationship between the 2 equations. The drop-off equation must give us the same value as we calculate for the density, when the length is 1 radii. To do that, the drop-off relationship must equate to 1 at the radius and to do that l must equal r. Which just states that the drop-off equation must equal the original value when there is no drop-off. We can't let it drop-off inside of the body because it is still building the density in there. So the drop-off can only happen outside of the body. Outside is where there is no input to the density. It is on its own out there. That is where it can drop-off.

So, whatever F1 and F2 stand for, it needs to be positioned at the surface. We want the complete body to create the density, but we only want it in a unit volume. We need a unit volume because we need to equate the two fields. To do that, we need them to be in a comparable volume. To create the density we need the complete mass of the body. To put it into a unit volume we need the density of the body. When we multiply them together, we get the right answer. That is heuristic, not a derived equation, but it works. It may not be perfect. It may not even give us the right value, but it works.

Maybe we can tease a bit more out of it, but I'm out of time right now.

Jared, we have dived pretty deep on this charge density issue, but it is related. I brought this in by asking whether the earth's charge field could extend out to the moon or not, which binding gravity seems to require and the charge-pause seems to refute. It is pertinent, but maybe not to the level we have been working lately. I apologize for that, but it is important and, more importantly for me, I am enjoying it.

Nevyn wrote:I brought this in by asking whether the earth's charge field could extend out to the moon or not, which binding gravity seems to require and the charge-pause seems to refute. It is pertinent, but maybe not to the level we have been working lately. I apologize for that, but it is important and, more importantly for me, I am enjoying it.

Yes indeed, I see your point now. I am totally fine with the conversation regardless and your previous post outlining the issue clarified it for me tremendously. Battle on, battle ON! I wish I knew how to help, but I'll keep working at it from my end too. Right now I've got a two-body charge simulation but we need a three-body for now. I'll try more stuff and see what happens.

I've misled you a bit. When I stated:

F1/(l/R1)^4 = F2/(l/R2)^4

That isn't quite correct. We can't use the same length in those drop-off parts. We have to use the distance from the body.

Let:
s = distance between both bodies
l = distance from body 2
Therefore:
F1/((s-l)/R1)^4 = F2/(l/R2)^4

You see, the distance used in the drop-off equation is not an absolute value either. It uses absolute units, but is actually relative to the body we are calculating for. It represents the distance from the body that is emitting the density. It is positionally relative, not size relative. It is actually the distance from the center of that body because the drop-off equation must equate to 1 at the surface of it. So l and s-l are different values, but they represent the same point. They just do it relative to the body in question. That is also why we need to know the distance between the bodies. This algorithm can be used to find the charge-pause at any distance, but it will only match the measured magneto-pause when we set that distance to be the relative position of the earth from the sun, for example.

Another way to look at this algorithm is that we are finding the charge density at some distance from each body. We are doing that in isolation from the other body, though. The earth has nothing to do with what the charge density of the sun is at some arbitrary distance. Only the sun controls that number, and the same for the earth or any other body. We can find those numbers without consulting the other body, so there is no requirement to use a common length or radius or anything else. We need to make sure that we do the same things to both bodies and that we have a comparable number at the end, but we don't need to consult the earth about the sun or vice-versa. Each side of the equation works in isolation to the other.

The only part that is shared between the two sides is the distance. We need that distance to represent the exact same point between the bodies at all times. But we also have to recognize that the equations are calculated from the perspective of the body in question. So we can't use the same exact value for that distance, we must translate it into the bodies coordinate system. That is why I subtract the distance from the full distance for the sun, but not for the earth. The distance is already expressed with respect to the earth (because I defined it as such) and the subtraction just gives us the side that is for the sun.

As an example, suppose our bodies are 100km apart. Therefore, s = 100km. The algorithm I used for my webpage uses a binary search algorithm to find the point of equality. That means that it starts in the middle, l=s/2, and calculates the charge density at that distance for each body.

Sun: Ms*Ds/((s-l)/Rs)^4 = Ms*Ds/((100-50)/Rs)^4
Earth: Me*De/(l/Re)^4 = Me*De/(50/Re)^4

It then compares those values. If they are equal, according to the tolerance set, then we are done. If the sun's charge is larger than the earth's, then we move half way towards the earth and try again. If the earth's charge is larger, then we move half way towards the sun. My actual algorithm is slightly different in that I keep halving the jump distance. So it will make jumps in s/2, s/4, s/8, s/16, etc. That keeps it inline with how a binary search algorithm is supposed to work.

If we assume that the sun has the larger field, then l = 50-25 = 25, and we try again.

Sun: Ms*Ds/((s-l)/Rs)^4 = Ms*Ds/((100-25)/Rs)^4
Earth: Me*De/(l/Re)^4 = Me*De/(25/Re)^4

Notice how the distance is being kept relative to each body, but the two distances will always sum to the distance between the bodies. The full distance must be the same for both sides, but the equality point needs to be able to move along that distance.

I'm completely following you here so far, Nevyn, and the math is about as transparent as possible. I'm not an amazing mathematician either but so far you're making sense and your rigorous variable assignments are impeccable. Please continue?

Meanwhile, I (painfully) finished the first chunk of my new charge model testing. This is just a proof-of-concept, but I'm finally able to use Rigid Body dynamics to demonstrate charge effects accurately. This is a very preliminary stab at the same problem, the sun-Earth-moon charge pause we're discussing. Take a look:

https://vimeo.com/327650447



This is my most advanced simulation of charge repulsion so far, in a 3-body problem as we attempt to address the sun/Earth/moon charge pause from every direction (eventually). In this one, all three bodies are identical in mass, radius, and charge emission. There is no initial velocity and no orbits involved here, this is just a test of the mechanism itself and the theory of charge repulsion.

As you can see, they repel each other well before a main-body collision. If I were to let it run long enough or increase the density OR velocity (or both) of the charge photons, we would find some sort of equilibrium after a certain amount of time. This is just a test of the simulation software itself, a proof-of-concept at best.

I'll start working on how to present this with accurate (relative) physics next. That is, the proper mass, radii, and distances involved. That part's pretty easy, but presenting it in a viewable format working with those distances is not so easy. This was a 3,000 frame simulation I squeezed down to 50 seconds (instead of 100, at 30fps) so it would play faster. I'll probably need a much, much longer simulation to work at the real distances but we've got nothing but time here.