Home Pyramidal Proton Great Pyramid 2nd & 3rd Pyramids Forum

The Pyramidal Proton Model

In a classic physics experiment, a magnet is encapsulated within a vacuum chamber. The magnet is shaped such that the two poles point towards one another but there is a significant gap between them. At one side of the vacuum chamber there is a pinhole in the wall. Atomic hydrogen gas, not molecular hydrogen, is shot through the pinhole in a thin stream. As the hydrogen approaches and enters the gap between the two poles of the magnet, the gas splits into two equal streams. One stream arcs over to the north pole and the other stream goes to the south pole. Refer to Figure 1:


The conventional explanation for the behavior is as follows: the splitting into two equal streams is not caused by the nucleus of hydrogen or by the orbiting of the electron. It is caused by the internal spin of the electron of each atom of hydrogen. This spin, and its resulting magnetic moment, are independent of the motion of the electron around the nucleus. The spin is an intrinsic property of the electron alone, and has nothing to do with its status in the hydrogen atom. It is decreed that one half of the electrons are north pole spin electrons. They are attracted to the north pole of the magnet. The other half of the electrons are south pole spin electrons, and they are attracted to the south pole.

However, if a thin stream of electrons is released in the same way as the atomic hydrogen, there is no splitting and deflection of the electron beam (experimental fact). It is explained that the charge property of the electron overwhelms its spin magnetic moment and thus no splitting occurs.

There are problems with the conventional explanation:
  • Are the electrons of the universe split into two, equal populations? Is this even split maintained, no matter what subset of electrons is involved?
  • There is no absolute set of coordinates in the universe. Your up is my down, your right is my left, and your north pole is my south pole.
  • If we have an intrinsic north pole spin electron, can’t we just flip it over and create a south pole electron?
  • Two simultaneous forces of attraction are required. The attraction of the electron to the proton and the attraction of the electron to a magnetic pole. What is an attractive force? This is action at a distance, which is in violation of a fundamental principle. Does the electron somehow “see” the magnetic pole and then attempt to go to it?
  • Since the electron is orbiting the proton, when does the magnetic attraction engage? Everywhere in the orbit? Any force has a vector.
  • Like charges repel one another, so why doesn’t a stream of electrons diverge to the two magnetic poles? How does electric charge repulsion in the electron stream overpower the magnetic attraction?
  • Hydrogen molecules can exist in two forms: “ortho” with parallel nuclear spin and “para” with anti-parallel nuclear spin. At 40 degrees Kelvin, 100% of hydrogen is “para”. As temperature rises, the portion of hydrogen which is “ortho” rises until it is 75% of all hydrogen at 275 degrees Kelvin. Apparently, proton spin is not intrinsic. Then why would electron spin be intrinsic?
  • When protons are present, there is a splitting of the atomic hydrogen beam. When protons are absent, there is no splitting of the electron beam. With which particle should we associate beam splitting?

Let’s start over with some fundamental principles, mostly from Meta Science:
  • Matter and energy cannot be created out of nothing or destroyed into nothing. There can be transformations of matter and energy constructs.
  • There are no attractive forces because there is no action at a distance. Any effect upon an entity must be immediate, local, adjacent.
  • Particles such as an electron or proton are not conscious. They don’t “see” anything. They don’t attempt to do anything. They are so “stupid” that they are unable to get out of the way!

Continue with theories from Meta Science:
  • The phenomenon of gravity is caused by a universal flux of gravitons, which push.
  • Light waves, i.e. electromagnetic waves travel through a medium, which consists of elysons. Much like waves through water.

End with some speculations:
  • I will propose that magnetism is a form of gravity, in which the graviton flux is focused or aligned. Specifically, there is a bidirectional “highway” flow of gravitons between the poles of a magnet. This traffic does not replace the random gravitational flux, but it “overlays” it.
  • The apparent phenomena of attractive force can be dissected and found to be a combination of repulsive forces and geometries.
  • In particular, the proton will have a geometry such that it strongly repels gravitons from one direction but only weakly repels them from the opposite direction.

Let’s begin with a proton having the shape of a right circular cone. The base will directly repel gravitons coming straight at it. The upper portion of the cone will slightly deflect gravitons coming from above. Refer to Figure 2:


In a random graviton flux, the proton will advance in the direction of its point. However, any sideways fluctuation of graviton flux will turn the direction of travel. The proton will wander. In a nebula of atomic hydrogen, constant collisions among protons will continually change the direction and speed of each proton. For reasons that will be explained later, the right circular cone proton will have its geometry altered in the following ways:
  • The cone will have four sides as a pyramid does.
  • The pyramid will be largely hollow at the base.
Let’s have a pyramidal proton (atomic hydrogen) approach the bi-directional graviton “highway” between the poles of a magnet. Refer to Figure 3:


Any proton entering the magnetic flux will have a random positioning which will ever so slightly point its tip towards either the north pole or south pole of the magnet. Once the proton begins to enter the bi-directional graviton “highway”, its shape and the graviton flux will turn the tip of the pyramid towards one of the two poles. Refer to Figure 4:


As the tip of the pyramid proton is aligned with one of the magnetic poles, it will be pushed, not pulled, towards that magnetic pole. There is no force of attraction. The gravitons hitting the base of the proton pyramid will forcefully push the proton in the direction of its tip. Gravitons, which hit the top of the pyramid, will only push weakly against the pyramid. All force is repulsive and in intimate contact with the proton.

It is proposed that the base of the pyramid has a parabolic concavity to it. The purpose is twofold:
  • This geometry accentuates the repulsion of gravitons at the base over a wider span of graviton flux from "below" the pyramidal proton.
  • It gives a position for an electron to be formed from Elysium. That is, a concentrated assembly of elysons.
Let’s consider the nature and behavior of the Elysium. The Elysium is a sort of ocean where the free path of elysons is extremely short. It is not free flowing like the graviton flux but acts like an atmosphere that has viscosity and therefore can have waves. The elysons are continually being pushed by the graviton flux. When they are pushed up against the surface of a proton, they pile up and “condense” to a liquid. The entire surface of the proton has a “liquid film” of elysons. On the outside of the pyramid, the graviton flux tends to push the Elysium film from the tip of the pyramid down to the base. This direction is “down” because most of the graviton flux comes from “above”. This localized effect is entirely due to the pyramid geometry. Once Elysium film has reached the bottom edge of the pyramid sides, it is exposed to the graviton flux coming from “below”. The film falls into the concave parabola and is pushed to the inward bottom of the concave parabola. As a liquid, it has no way out, but it is continually being pounded by the graviton flux. The gravitons lose momentum to the Elysium liquid. That is, it rises in temperature; it heats up. A bubble of vapor Elysium will begin to form at the base of the parabola. The bubble expands and an electromagnetic wave - the surface of the bubble - forms. This wave goes forth out into the open Elysium. Depending on the frequency, it will be heat wave or a visible light wave, etc. The thickness of the Elysium liquid film is exaggerated in Figure 5 so that it is obvious to the reader.


There are three important factors about this mechanism shown in Figure 5:
  • Every step of the process is actuated by repulsive force. There is no attraction.
  • The process is irreversible. All progress by the liquid Elysium is downhill.
  • The process is perpetual; it never stops.
Suppose that the pyramidal proton impacts against another proton. A sudden impact would reactively empty the concave parabola of its liquid Elysium. This would be the ionization of an electron: a liquid mass of elysons. This electron ball of liquid Elysium would go out into open space where it would eventually “evaporate” into vapor Elysium. One could envision the Sun as a huge sphere of liquid Elysium, aside from the fact that it has protons, neutrons, etc.

Creation of Deuterium

The claim is that an electron consists of a formulation of elyson particles just below the base. The graviton wind and the geometry of the pyramid base lead to the formation of a higher density, higher “pressure” cloud (perhaps liquid) of elysium particles. This high density Elysium presents a highly repulsive effect, which fans out like a searchlight. In contrast, the tip of the pyramid is the least repulsive position on the proton. On a relative basis, it is “attractive”. The sloped sides of the pyramid have a moderately repulsive effect. With my prose, a picture is worth a million words. Refer to Figure 5 for the appearance of atomic hydrogen:


The pyramid is blue, its base is red and the “electron” is pink.

A nebula of pure hydrogen is compressed by the gravitational flux. By means of this gravitational pressure, the temperature of the hydrogen gas is raised to several million degrees Kelvin. At this temperature, it is claimed that nuclear fusion begins, with atomic hydrogen fusing in three sequential collisions into helium. This is the point where this model parts company with conventional nuclear fusion. Instead, it is proposed that two atomic hydrogen protons come together, base to base. This juncture is not planned or predestined. It is a random occurrence that happens perhaps once out of every trillion collisions. As the two protons approach one another, the repulsive effect of the two electrons (elyson “exhausts”) push against one another and are diverted outward in a direction which is orthogonal to the axis of collision. When the two bases reach the point of contact, the repulsive electron “exhausts” have been partially encapsulated and partially dispersed. The bases contact one another much like two spacecraft docking in space. Refer to Figures 7 through 11:






Two very important results are achieved:
  • There has been no energy gain to the local environment. Some of the relative momentum of the two protons has been transferred to the surrounding of elysons.
  • Some elysons have been encapsulated within the two hemispherical concavities of the protons. This trapped elysium is at the temperature of the local environment: several million degrees. But the surrounding Elysium is also at several million degrees.

In conventional nuclear physics, the two protons have collided at a relative velocity of about 1,000 miles per second. How do they stay fused? Oh, yes, the strong nuclear attractive force, which appears out of nowhere. This strong nuclear force binds the protons together but not too close, because, yes, there is a new, repulsive force, which holds the protons just a little bit apart. In this model, there are no attractive forces; there are no magic forces that appear out of nowhere.

With the base to base union complete, we have deuterium. Let’s compare deuterium to molecular hydrogen, which is shown in Figure 12:


The chemical bond between the two hydrogen atoms is pyramid tip to pyramid tip. There is no force of attraction. It is a position of least repulsion by the graviton flux and the elysium. No further bonding will occur because of the dominant repulsion by the two electrons.


In comparison, the deuterium has two pyramid points and eight sloped sides. There are no repulsive bases. On a relative basis, the deuterium particles are not repulsive. Given that they have twice the mass of atomic hydrogen, and little repulsion, the force of gravity will cause them to migrate “downward” and to huddle together. Deep enough within a star, the high gravitational pressure will begin to force the deuterium nuclei to bond together, face to face. A small amount of elsium will be trapped between any two faces which close together. This trapping is the weak nuclear force. The trapped elysium is not attractive but weakly repulsive. As the deuterium binds together, a crystal structure will build. Because there is still an elevated temperature, the crystal growth will be random, disordered. There will be many voids within the crystal. As the crystal grows, voids in the center of the crystal will be diminished under the force of crushing. However, they will not be reduced to zero. Consider that the condensed deuterium will lead to a higher gravitational gradient through the crystal. As the crystal grows further, eventually the gravitational flux in the center of the crystal will be greatly diminished.

What holds deuterium together? There are two factors: the high inertial mass of the two protons and the “pressure” of the gravitational flux. The elysium inside cannot “get the jump” on the gravitational flux.

Hydrogen Bomb

The reader may state that the hydrogen bomb is a clear case of fusion releasing an incredible amount of energy. But does it? The reaction is actually between tritium and deuterium. Refer to Figure 13 for the arrangement of tritium:


The tritium has one electron exposed and one, fully exposed, pyramid tip. This would be its chemical bonding point.

Let’s look at a possible collision between tritium and deuterium. Remember that the collision will not occur in the zone of the exposed electron. Refer to Figures 14 through 18:






The release of the “100 million” degree elysium is apparent when the surrounding elysium is at typical, ambient temperature (~70 F). If this reaction occurred in the core of a star, it would have no apparent effect.

Pons-Fleischman “Cold Fusion”

There was the occasion of the Pons Fleischman “cold fusion” experiment. They used a palladium lattice (fine structure) to adsorb deuterium and then applied an electrical “trigger” to initiate the process. The water also contained significant deuterium oxide. Results were an excess of energy, helium, tritium and neutrons. The entire scientific community debunked their claim. After all, it takes “100 million” degrees to initiate fusion. We all know that from fusion reactor experimentation. However, it is possible that the following happened, as shown in Figures 19 through 26:









In this particular collision, tritium is made by fusion and a neutron is released. Such a collision is random in its results. Helium could have been formed or two neutrons could have been released. Since the overall reaction spreads through the deuterium, it is far more likely that neutrons are released. There would be very little helium formed and a modest level of tritium formed.

It is proposed that Pons Fleischman achieved nuclear fission of deuterium. High temperature on a macro scale was not and is not necessary. With proper orientation of deuterium and a trigger, the release of energy can be obtained on a very small, controlled scale. The experiment was dismissed out of hand because traditional scientific theory was allowed to over rule actual observational data.

A Solar System

Let’s begin with a large hydrogen nebula at time “zero”. It is irrelevant whether the hydrogen is molecular or atomic. The gravitational flux will begin to compress the hydrogen inward. By means of collisions, the hydrogen will rise in temperature (velocity) and it will become atomic hydrogen even if it began as molecular hydrogen. Since the model assumes that a proton (hydrogen) is a pyramid with a concave base, the primary means by which hydrogen atoms will increase in temperature (velocity) is from gravitons hitting and repelling from the proton base. Eventually the temperature of the hydrogen cloud will reach “several million” degrees where fusion can begin. Even conventional science agrees that the force of gravity must be the source of energy bringing the cloud to the point of nuclear fusion. When the cloud has reached a concentration and temperature to initiate fusion, this model predicts the following:
  • The fusing of protons will “remove” (i.e. encapsulate) “100 million” degree elysium from the hot cloud. Thus, the fusion reaction stabilizes the temperature of the cloud by hiding high temperature energy.
  • Because the newly formed deuterium has no electrical (coulomb) repulsion, it will not strongly repulse other nuclei. The deuterium will “sink” and begin to crystallize under the gravitational pressure of the entire cloud. The beginning of crystallization will promote itself, because the gravitational gradient between open cloud and the crystal will rapidly increase. The cloud as a whole will shrink more rapidly.
  • The crystal structure is not a perfect solid. There are numerous, highly contorted voids. When the crystal has grown large enough, the gravitational flux within the center of the crystal will diminish greatly. Since it is the gravitational flux that holds deuterium together, and since the elysium within the voids in the crystal will no longer be at “100 million” degrees that it is out where fusion is taking place, the deuterium will begin to explode in the center of the crystal.
  • The released elysium will expand rapidly through the voids of the crystal. Such a shock wave will promote more “popping” of deuterium. The initial release of elysium will self promote its further release.
  • The shock wave of the elysium will propagate through the crystal up to the velocity of light. The crystal will blow apart. However, the overwhelming “popping” of deuterium will quickly fail for a very simple reason: gravitons move much faster than light. They will rush in through the voids, apply “pressure” to the remaining deuterium and stop the “strong” fission process. However, the “weak” fission process will continue: the breaking apart of many of the side-to-side adhesions of proton pyramids. Thus, all the isomers of all the isotopes of all the elements will be born simultaneously in a single explosion.

Nuclide Chart

A picture of “all” nuclides is shown below:

The one outstanding feature of this chart is that it is not a Plan, not a Design. It is random. All the nuclides are simply fragments of the polydeuterium explosion. It would be more accurate to name it the Fragment Chart. Both stable and unstable nuclides are formed. It is critical to understand that, when the inflowing gravitons put a stop to the “strong” nuclear fission, the outer portion of the crystal will break apart along the voids within the crystal. This gives birth to nuclei larger than deuterium, with open pyramid bases being a rarity on these nuclei. This point cannot be overstressed.

Early in a star’s life, the basic fusion reaction that creates deuterium, the forming of polydeuterium balls, and their exploding apart - all would happen at a great rate. Thus the material for planets would be formed very quickly. The star would continue to shrink and planets would form. Over a considerable period of time, this process would decline for the following reasons:
  • The disintegration of polydeuterium balls would produce a growing population of light elements, particularly helium, which have high coulomb repulsion. These elements would interfere with the fusion of hydrogen, which makes deuterium. In effect, they would “poison” the process.
  • A drop in this fusion process would lower the removal rate of hot elysium from the star. In consequence, the incoming energy from the graviton flux would raise the temperature of the star.
  • In sharp contrast to conventional nuclear fusion theory, a higher temperature would cause a higher failure rate of the collision of two hydrogen atoms to form deuterium. The reactants would have too much velocity and would break apart at the instant of collision. In chemistry, many vapor phase reactions have an optimum temperature for success. A higher or lower temperature will reduce the success rate of the vapor reaction. A classic case is the oxidation of nitric oxide. It progresses best at about 21 C. Higher and lower temperature diminishes the reaction rate.

We can see today that our Sun has a diminished rate of fusion production. Sunspots represent the formation of polydeuterium balls and huge flares represent the explosion of such balls. However, there is no apparent ejection of heavier elements out of the Sun.

Radioactive Decay

What is radioactive decay? An unstable isotope does not give radioactive decay continuously. Instead, it is an event, which occurs to the unstable nucleus sometime in the future. There is no clock ticking away in the nucleus. Instead, the nucleus experiences a collision with perhaps a neutron, or a deuterium atom or whatever. Why does the collision happen and why is it successful in causing decay? Let’s look at a brand new nucleus that was formed in a polydeuterium ball explosion as shown in Figure 27:


As can be seen from the figure, this nucleus does not have any exposed proton bases. That is, there are no exposed electrons. In essence, the nucleus has no repulsive “protection” from collisions. The pyramid points and faces are exposed to any collision with a foreign body. Suppose that a neutron impacts with the nucleus. This collision may cause a proton within the nucleus to be tipped over or even broken free of the nucleus. A result of the collision is the sudden exposure of a proton base upon the nucleus. That newly exposed base will form up an electron which has a repulsive coulomb force. In this portion of the nucleus, there will protection, by the electron, against further collision. With enough successive collisions, enough proton bases will be exposed to protect the entire, remaining nucleus from any more nuclear collisions. The nucleus will now be stable.


The stable nucleus’s chemical behavior will now be determined by the number of pyramid tips that are exposed and available for chemical bonding. This bonding will be pyramid tip to pyramid tip. We might examine the nucleus and determine two important things: the total number of protons and the total number of available pyramid tips for chemical bonding. We will then decide that it is oxygen, or nitrogen, or whatever. We will give it a position in the Periodic Table. However, this nucleus will have absolutely no idea or awareness of what isotope and element name we have assigned to it!

Conventional Chemistry and the Periodic Table

It has become a long tradition and a deep belief that chemical bonds between atoms are directly related the number of outer electrons, which can bond with one another. However, this model presents the case that exposed electrons are entirely repulsive and they tend to keep atoms separated. The chemical bonds are actually achieved by the exposed pyramid tips on the nucleus of each atom. This bond is not an attractive force; it is a geometric positioning which has the least repulsion between the two atoms. The atoms do not attempt to bond together; they are pushed into “bonding” by the surrounding elysium.

The Periodic Table is a human invention. Reality could care less. This is a ritualistic system, which tells us the weight of a nucleus and how many bonds it can form, but it does not explain how or why such bonds form. The Periodic Table gives the false impression that there is a unique orderliness to the array of elements. In truth, the elements and their isotopes and the isomers of the isotopes are created randomly in the disintegration of a polydeuterium ball.

Let’s consider nitrogen. It can have as many as five bonds yet it may form very stable molecules with less than five bonds. Nitric acid, HNO3, has all five bonds but ammonia, NH3, uses only three bonds. Why? It is a matter of geometry and electron repulsion. In this model, the hydrogen atom is a single pyramidal proton. It can form a chemical bond with its pyramid tip but has a base with wide electron repulsion. Suppose that pyramid tips on the nitrogen nucleus are in groups, or closely aligned. Once a hydrogen atom connects to one pyramid tip, its electron base repulsion may isolate or “smother” another pyramid tip on the nitrogen nucleus thus stopping a chemical bond. So, perhaps the three hydrogen atoms in ammonia effectively place the other two bonding points out of reach for further bonding. In nitric acid, perhaps the two bonding tips on an oxygen nucleus line up well with two bonding tips on the nitrogen nucleus, thus leaving three more bonding points on nitrogen to bond with another oxygen atom and a hydroxyl group. Geometry determines what will happen.

Examine a neon nucleus. It would have enough exposed electrons such that no position on its nucleus would be available for chemical bonding. Not only is it safe from nuclear collision, it is safe from chemical attachment. It is quite interesting that there are three, stable neon isotopes, Ne21, Ne22, Ne23. One wonders how many isomers there are for each isotope.


The proposition that a proton is a pyramid with a concave base, leads to explanations for nuclear fusion, nuclear fission, the forming of planetary material, solar and planetary explosions, the creation of all isotopes and isomers, radioactive decay, nuclear stability and chemical bonding behavior.

But is the proposition true? There is a test for establishing or rejecting the proposition. The vacuum chamber with the magnet inside would have to be assembled. With adequate temperature, atomic hydrogen would be injected into the vacuum chamber to pass between the poles of the magnet. If the hydrogen stream splits into two streams, one to each magnetic pole, then it would be established that the apparatus works correctly. One would now, in a second trial, inject atomic deuterium into the vacuum. If the deuterium behaves like the hydrogen, then the proposition of this paper is false. If the behavior of the deuterium is distinctly different, then there is evidence for further consideration of the proposition of this paper.

A Final Note

All of the bizarre properties and behavior of Helium-3 is predicted by this pyramidal proton theory. Hint: it has three protons, no neutrons. But this is another story.