A Non Fiction Trilogy

CHAPTER 5 ELEMENTS

June 9,2018

This chapter is another big pho-par, and this one has not been re-done yet, it will need to be as chapters 3-4.5 have been. It should be more in line with the thoughts expressed in red below. This chapter was done 25-30 years ago, while very excited about things, and quite obviously now I see its stupid, no way to have linear motion of N to N pairs if they are spinning around each other as described below (after new red section)!

Because I don't know when I might kick the bucket, I am including here ideas that are very preliminary.

With regards the accretion of a Nucleon. Originally I had a set of two rows of 137 PP turned 90 degree from each other topped with a single PP on each side also turned 90 degree to that face.

But as of 2017 and going over chapter 2-5 in finer detail, I have a problem in getting such a N to form. It appears, though I am still working on the details, that I might end up with, at best, one complete row of P's on one side, but only a half row on the other. This is because in order for the particles to line up at center of mass (com) they must rotate to that point. And scenarios to do this do not allow a second full row. To get that on the second 1/2 row on one side P would have rotate off the end P of the other row and land on top of a established row, which would prevent center of mass alignment.

May 25 2018

Looking at chapter 4.5 and the above. ANY particle that ends up unable to re-achieve linear motion in total cannot fit with the dynamics (system) of chapter 6. So the above, full one side, half full the other face, would not work, unless I have missed something on how linear motion can be achieved. And since can not get two full faces, only one full face then can be a N.  Now to fit them together as compounds.

June 6

Reading about Neutrons, wow how fascinating as regards my theory. This is where the rubber meets the road.    In standard physics the perception of a Neutron may be way off from reality.  Furthermore, the idea that a electron does not decay (at least for 10 to the 28th power years, is polar opposite of my hypothesis were it probably is formed and destroyed millions of times a second.   And why aren’t the energy states, including emitting a photon considered a type of decay, definitely is a change?

But I can’t figure neutrons out via my hypothesis, they would have to still be able to be accelerated, but not have a capacity to discharge a PP like a proton. Which should be possible to arrange with my N as so far derived from chapters 2-5. But need to figure it out, if possible.

June 27

As regards linear motion of the 4P case, with particles all balanced  and A-B at midpoints yes then the force of rotation off C and D could accelerate B linear, but is this the real mechanism to accelerate  C and D linear? Or should all motion ACD go into B and shoot it off like a cannonball? Well its because there is equal rotation off C and D so this restores linear in A (like principle 12), whereas it had been rotating because of lack of balance.  One might say that the equal distance C to end and D to end, combined with the contact CBD create the balance.  So then if B is removed would ACD still move linear, assuming it appeared at that state from a rotation (though not possible). Indeed it would seems yes, therefore Principle 18 needs to be rethought     So it could be said when A cannot rotate it pushes anything linear anywhere along it, its only when A can break into a rotation that linear transfer is broken.

So how then should I re-visit Principle 18. For contrary motions there is already rebound of sorts. For dual torque  and dual torque and rotation there could be rebound. But dual rotation may instead cause linear acceleration. After all this keeps the direction of motion more as it where in the rotation at said point in time. Further the only way now to accelerate a N is by the process of an overtaking hit on an extra and rotation to a midpoint, about a one in 1000 chance, acceptable, but with dual rotation acceleration that becomes much greater.

June 29

Further this possibly is the only way (dual rotation = linear) to now get two N together. To either form a "true" N or to form elements.

So appears from chapters up to 5 I am left with either a full row and an "extra" (pusher P). Or a lopsided 1/2 row and extra. Or possible a full row and 1/2 combo with extra. So those are, or the pieces to use to form,  a N (Nucleon) From there must have a way to have them come together, perhaps somewhat like in the rest of the chapter as was, and to have a distinguishing characteristic between Protons and Neutrons. And it would be nice to form then in curls still, as this made a match to the periodic table, as well as leaving Carbon as a unique arrangement where it had the first arrangement with 2 places on each side (4 sides) with 2 contact points, to have PP interact with.

July 10

So if this hypothesis is correct it would have to explain the periodicity of the periodic table. which, by the nature of the hypothesis, playing with tinker toys as it where, that is very straight forward and geometrical. No other explanation comes to mind that makes any sense, other than what I had already figured before, with the curls (as below, old part of chapter). These curls in following patterns that repeat, creating  sets of contact points on the sides of the N's that make up the elements. They provide by similar contact point which create similar rebound or such interaction from PP hitting on sides, this is therefore what creates the similar properties, the periodicity of the elements.

For this to happen then the extras have to be, on each face, running in opposite directions. This cannot be had without two back to back faces per N. But as explained above the most I can get is one full face and backed up by a half face. This would be suffices to have the extras perpendicular to each other, but this scenario would only produce spinning N, at least over the long run. Unless perhaps with a change of the dual rotation principle to induce linear motion again, as described above. Then perhaps this would all work, so must think about that.

July 25th

An other issue is the extras, I apparently had not thought it though before, as the extras would need to be at center of mass in order the transfer motion as a unit with the faces. This can only be achieved having only two half faces back to back, then an extra would be able to rotate, or be rotated around, in two axis'. So this leaves the TWO HALF ROWS WITH AN EXTRA ON EACH FACE OF THOSE ROWS. This then should transfer linear and rotational motion as a unit, including re-rotation which reverses direction of unit (N), though it would be complicated and after principles 19.4 and 19.5 eventually.

August 2-Sept

Some interesting thoughts. The possibility of dual rotation = linear is really very much like linear motion of a doublet via principle 12. And when the arms are not in balance rotation occurs off the longer arm with more force to it, in effect, re-directing all the motion in that predominate direction. Similarly on a see saw, the heavier side causes all the see-saw to swing that way, although at varying speeds depending on the balance, as the force of gravity is in play. However on the primary level here the particles themselves are the carriers of gravity, by intrinsic inward motion which is part of the overall flow of the system, but not effected on a moment to moment basis, so the speed of rotation is entirely proportion to particle speed, regardless of balance issues.  Thought the longer the arm the slower the apparent motion because of more area swept, but this is not the same thing I was talking about.  So this also implies that in the system motion (inward component) which shows up on the slightly more macro analysis of system motions, not the flux by flux motion, but the more overall configurations, there must be the math for the gravitation, something that surprises me, as I had not thought gravity could be unified with electricity, but perhaps it can be in a way.

I will try then to figure the transfer of motion though a N, as above it needs to be one with two 1/2 faces and 2 extras. Using dual rotation equals linear, this would also imply that transfer of motion of a PP across a face is from all points of contact, as in falling on the face the chance of mdpt contact of the PP is very unlikely (as per principle 10). That is when a dual rotation situation of any type (two or more points of contact) is in play, motion is no longer transferred from mdpt contact only, but from all points, this itself implying then that, in principle,  linear motion occurs when no rotation is possible.

Lets consider then an overtaking hit, this PP overtakes the N and hits on the edge of the face in its path. Leaving out and hits then that contact the extra for now, it would, as in cases in chapters 2-4.5, "establish" a unision motion, and any excess motion above that would go into torqueing. after torqueing it would then rotate and land on the face. I have left out the details for the moment, but when contact is made across the face it will be, I assume for now, squared up to the face (because that's what the torqueing did). Here it would have some overhang "above and below" the face, assuming it hit first on a PP at the top or bottom of the face. That is to say if it hit on one of the middle PP of the face first, when falling on the face it would overhang past its mdpt and therefore continue to rotate around the face, not having a dual rotate quality, that is not having contact pts on each side of its mdpt.

So then back to falling across the whole face, it's mdpt then would fall within the area of the face, causing dual rotation, and as now considered, would accelerate the face from all points of contact. Here all rods would rotate off the PP as well as rotating the extra, but first this would cause the rods to hit the perpendicular rod at the bottom (or top) of the other face, casusing instead re-rotation onto that face and contacts and linear motion of the first face into the second. Thence accelerating both faces and the far extra linear. That is locked mdpt contact and contact pts all on one side of the mdpt act as if dual roation is occurring. The principle here I need to deduce better though.

So this leaves the first extra left behind at its old speed and the rest of the N moves away. That first extra then, being at a speed much less than the PP flow has energy as so described in chapter 6.

August 30-Sept

Now considering a direct hit, of the same type (not hitting the extra) this might work as follows. A PP hits one of the outer PP on the face all motion of those two PP is stopped and they torque. All other N particles on the face and the extra keep going. But the PP's on the back face and the back extra are also held by the collision and, if it was the particle at midpoint that was held, transfer all their motion to that particle which accelerated its torque, but said torque of P hits P's below it transferring all motion downward and at a perpendicular. But these same particle in moving forward causing a combined motion and new vector, resulting always in hitting the extra. This thereby  causes it to rotate (off all particles except the midpoint P which is held, therefore rotation is possible, virtually  over the instant) and then dual rotation and then linear, accelerating all these particles in a opposite direction to the current linear motion. That combined with the downward component causes a new vector, directed "backward into the heart of the N which causes all motion to flip to the perpendncular via contact pts with back face causes all motion to revert to linear in the opposite diredtion to orginial motion (again).

This then is the N reversing and the PP that was a direct hit turning into an overtaking hit.

If there is an plastered particles on the back face the rest of the N has to rotate on it until its at center of mass of the face, then it can all revert to linear, including the new "extra" , even though it is not com itself up and down.

Well its more complex than that, and don't think it would work like that,  because there is the reverse of the linear acceleration here, that is multiple points of acceleration on each side of the midpoint should cause acceleration as the reverse of dual rotation sort of thing. So the face would rotate but it is not held as it does accelerate the plastered particle.  Now if a PP hit (direct hit) on that plastered particle then there would rotation of the face. so indeed there is full reversal of all particles on N and plastered particles too.

Need to make principles this all getting too confusing;

Principle 18 and 19.6 need revision. Perhaps 15.6 and 16. But Principles 1-14 and 17 are okay, but maybe for only 2 or 3 rods interacting. Over that it might be different. So

1. Dual rotation equals linear motion

2. What happens with accelearion in the opposite direction?

3. what happens when a face rotates into another face, s motion passed from or too all points. or only midpoints?

etc.

Figure 4.3 is the relevant figure.  Here when C and B, or any number of particles. are  on opposite sides of the midpoint of A, the senarion needs to be changed to one of a result of linear motion resulting of C & B. Likewise if C and B where to hit A simultaneously, it would accelerate it linear. The same with any number of P regardless of there positioning, as I see it.  Only when a P is at the midpoint of A does the situation change and orbiting occurs. Any number of P touching this midpoint P on the loner arm can be orbited together by Principle  17 .

Now then for the case of a  (1/2) face (Af) rotating into the other face (Bf) in a N. If there are two P's at perpendicular, both below the midpoint of Af  it  would rotate off the outer particle. If one is above and one below midpoint, then the face Af would undergo dual rotation on those particles and result in linear acceleration of all. If one P is at midpoint on Af then it would be orbited (swept up) by the rotation of Af. If there is a P between the two P's then Af would rotate off that particle, still orbiting the midpoint P.

If a series of P's are adjutant to the P at midpoint, then they all are orbited as like Chapter 3 case 10C. However the base P being rotated on by Af, now here considering it is rotating into Bf (a full face-1/2 face as it is now) the base P cannot be moved if it is being held by a direct hit from another P.  This being the case all motion breaks back to linear and one could suppose it accelerates all the other P on face Bf and the extra there in linear motion. But resorting to principle 5, the least change in the system, sense there is a battle between Af rotating or being linear via Bf, is for that interaction Af to Bf to lock up and all motion drains into the extra on Bf, accelerating it at high speed.  The high speed "electron" so needed to produce light. So this is a good scenario to go with than, unless it is found to seems  to violate consistent logic as to motion principles.

HOWEVER THEN CHAPTERS 2-5 NEED TO BE REDONE TO ACCOUNT NOW FOR THE DUAL ROTATION LINEAR ACCELERATION AND HOW THAT WOULD EFFECT THE BUILDING OF THE N!!

Now further consideration needs to be made of overtaking and direct hits that do imping apoun the extras. As well as hits on inner PP on the faces.

Now further consideration needs to be made of overtaking and direct hits that do imping upon the extras. As well as hits on inner PP on the faces.

Sept 26

(New) Principle 18: For a System of three rods, when a rod undergoes dual rotation it will accelerate the rods with a linear motion.

Principle 18.1 Those rods can only receive and achieve linear acceleration if the contact is at their midpoints. Else they re-rotate around the impacting rod.

This is necessary for any consistent principles. That is if Principle 12 occurs, then also if the reverse directional of saying if a rod A hit rod B (stationary and perpendicular) from its midpoint, if rod B was accelerated linear then Principles 12 and 13 would be in-validated. But B cannot rotate forward as what would it rotate on, or if it did it would immediately impact A in the shorter arm going backward, and re-rote off A.   So, either this is what happens, or all motions goes in all directions and follows, as it were the path of least resistance, adding to that the flow, in addition to going in all directions is also predominate, as strange as that may seem. So, with one of these or elsewise there may be an overriding principle for motion.

So let’s revisit the issue of torque, because I need to know more about motion within mass.      If you consider the impacting particle alone, and the motion drawn to the point of impact its all symmetrical unto itself. Relative to the impacted particle then, torque would only result if the window of contact restrained the motion in some way, and that by needs would have to be related to the shape of the particles.

One can say this: the slice of the impacted particle with the least breath is the slice (as related to the ends of the impacting particle) is the slice conducive to perpendicular alignment.

However consider too, rather that all motion vector ankylosis only, the friction of any point from the surrounding points, or area is infinite. So the instantons transfer of motion within mass area is as much a “pulling   along of adjacent point as it is a question of vector shifting. This pressure gradient is directly related to the primary vector motion of the particle. If it is lengthwise the pressure of pulling is from and too the short axis of the rod. If, as postulated, it is primarily perpendicular to the ends when created by God, then the pressure is in proportion to the long axis. When it then crosses the window of contact NO NO this would not lead to anything but parallel alignment.

So for orbiting, this means the notion of contact across the window in a straight line can be discarded if needed, as linear motion is no longer dependent on that alone, but on the symmetry issue. So one might then consider any angle is possible across the window of contact, short of parallel to the rods.

So again, for orbiting that might leave us with the notion that rotational motion in the rod is like linear motion with a continually changing angle of direction. In essence the same as the discussion on infinite numbers in __________, here every rotation instant has a linear motion, the next instant a certain different one, but there are no two successive angles so as to say it changes from this “angle to that” only the rate of change over any interval of rotation, determined by the rotation precession of on rod on the other. So that at 90 degree turn there is a precise angle change determined not by speeds, but by fixed geometry.   I would also say, similar to Newtons first Law of Motion. This angle change is “known” in any instant, such that it is passed across the window of contact, causing as it where a rotation/orbiting of the next rod as opposed to linear acceleration. That is unless acted upon by another force/impact it continues as is “imprinted” in the motion of the rod (however as it precesses that is a continually changing imprinting also!

This also means there is no need for the midpoint to transfer rotational motion. In the cause of dual rotation then than linear the longer arm should orbit the 3rd rod (if its balanced at midpoint, otherwise it re-rotates back around the other rod). Only when it reaches a point of equal arm lengths does it produce linear motion. However on the face of a N it is a case of all rods in contact and the rods rotated off is forced back into the rods next to it stopping rotation so by mutual contact motion seeks its own distribution of balance

That is principle #   is out on the face of it, it is the symmetry that counts, but leads to de-facto that principle working in most cases. So if 3P and BC on opposite sides of A’s midpoint and rotate to symmetry, linear motion is achieved. If C is at midpoint and have 3P case of BC together, the rotation is the same around point on B because no rotation off of C if it is at midpoint of A.   if C was past midpoint and they were tighter, either would not come together, or is all question of distances from midpoint as to any balance being achieved or just spin. Etc. stuff if follows cases here.

When there is a series of P together and P across there is feedback so that re-r goes linear. If is space between them is no feedback and pre-dominate rotation occurs. So for the 4P case balance at midpoints is not necessary as such, as feedback + linear.     So need to put this all into concrete wording/principles.

hen also expound on

3P case a dual rotation principles

Principle for two rods orbited together

Principles for 3 or more rods in succession being pushed by a perpendicular rod.

Old Chapter

If two nucleons collide they will usually “rebound” from each other.   But if they came together as Fig. 5.1, and if rotating perpendicular to each other, they would “spin” around each other as all motion of A goes through Ae, pushing B, causing B to rotate around Ae, all motion of B goes through Be, pushing A, causing A to rotate on Be, etc.

Figure 5-1

Proceeding further:

Principle 19:     Each successive nucleon always turns in a direction toward the center of mass in stable elements.

4 nucleons = Helium (approximate atomic masses), He being two more nucleons added on to the set in Fig. 5.1.   A side diagram of He would be

He – side view

No outer extras shown

Shaded is back a bit relative to middle extra

O – extras

/// - extra back from page

1:6 Width to Length Ratio

Figure 5-2

The “extras” on each side of a nucleon are in different (perpendicular) directions, so as nucleons “accrete” they curl around.

Off C & D as in figure 5.2

off C                                              off D

Figure 5.3

After He they continue in the same sequence per principle 19, one set of four off each end gives a complete curl for each, ending up with a 12 nucleons or Carbon 12 on the periodic table

Numbers represent (approximate) atomic masses as each element is formed.   Extras shown with same width as proto-nucleon (proto-nucleon should be 2x the extra’s width).

There are 2 primary things to be noted.   One is the end nucleons face's with their extras (one for each nucleon), which present a spot for contact with PP’s.   They may have one of 4 combinations, as noted on chart 5-1.   Also as each new nucleon is added the sides, being all squared up, present points of contact for any PP hitting on a side (there are a total of 4 sides).

The actual hit and interaction of PP with the ends has not been worked out but should begin to explain magnetism.

The PP hitting the sides “interacts” basically as described in Chapter 4.   This has not been worked out either.   Suffice to say here that where the midpoint of the PP lands and having the contact points equal or unequal in number and area on each side of the midpoint is what determines much of the interaction.   If the midpoint contacts a contact point then by Principle 10 the PP is always unbalanced.   If the midpoint falls in a “gap” and also the number of contact points on each side are equal, then the PP maybe balanced.

Carbon 12 (side view)

Top View

Figure 5.4

There are 2 types of contact points, the outer “wings” (row A,D , Figure. 5.4) which are always solitary on the right or left; or the central section (row B,C , Figure. 5.4) which is “doubled up” (for each set of nucleons).

There are two types of gaps, the larger gap between successive repeat positions in each complete curl, and the small gap between two adjacent nucleons (in between the doubled up contact area).

d.

CHART 5.1 -   ORIENTATION OF THE NUCLEI,   PROTON - 36Ar

The following is a chart of each element up to Ar36 with the number of contact points and gaps listed, wg is the widegap, sg is the small gaps where nucleons pres on each other. Rows A, B, C, and D are as in figure 5.4. For the top and bottom count the rows as looking at the top for the top rows, and as looking from the top through to the bottom.

For the right and left sides count the rows as looking at the right side for the right side and for the left side as looking from the right side through to the left side.

The ends have to do with the orientation of the extras on the end nucleon of each nucleus (11 and 12 in figure 5.4).   This orientation is described in the first four boxes on the chart, after which they repeat because of the “curling” of the nucleus as successive nucleons are added (as shown in figure 5.3).

wg   CB           sg   A   D                     Ends

 1   Proton Top Bottom Right side Left side 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 One vertical, one horizontal.    Same level.    Same column. 2   Deuterium Top Bottom Right side Left side 0 0 0 0 0 2 1 1 1 1 0 0 0 1 0 0 0 1 1 1 Both horizontal.   Same level.   Opposite columns. 3 Top Bottom Right side Left side 0 0 0 0 1 2 2 1 - 1 1 - 0 1 1 0 1 1 1 1 One vertical, one horizontal.   Different levels.   Opposite columns. 4   Helium Top Bottom Right side Left side 1 0 0 0 2 2 2 2 - 1 1 1 1 1 1 1 1 1 1 1 Both vertical.   Same level.   Opposite columns. 5   Out Top Bottom Right side Left side 1 0 0 1 3 2 2 3 1 1 1 1 2 1 1 2 1 1 1 1 One vertical, one horizontal.   Same level.   Same column. 6   6 Li Top Bottom Right side Left side 1 0 1 1 4 2 3 3 2 1 1 1 2 1 2 2 2 1 1 1 Both horizontal.   Same level.   Opposite columns. 7   7 Li Top Bottom Right side Left side 1 1 1 1 4 3 3 4 2 1 1 2 2 1 1 2 2 1 1 2 One vertical, one horizontal.   Different levels.   Opposite columns. 8   Out Top Bottom Right side Left side 1 2 1 1 4 4 4 4 2 1 2 2 2 1 2 2 2 2 2 2 Both vertical.   Same level.   Opposite columns. 9   Be Top Bottom Right side Left side 1 2 2 1 4 5 5 4 2 2 2 2 2 2 2 2 2 3 3 2 One vertical, one horizontal.   Same level.   Same column. 10   10 B Top Bottom Right side Left side 1 2 2 2 4 6 5 5 2 3 2 2 2 3 2 2 2 3 3 3 Both horizontal.   Same level.   Opposite columns. 11   11 B Top Bottom Right side Left side 2 2 2 2 5 6 6 5 2 3 3 2 2 3 3 2 3 3 3 3 One vertical, one horizontal.   Different levels.   Opposite columns. 12   12 C Top Bottom Right side Left side 3 2 2 2 6 6 6 6 2 3 3 3 3 3 3 3 3 3 3 3 Both vertical.   Same level.   Opposite columns. 13   13 C Top Bottom Right side Left side 3 2 2 3 7 6 6 7 3 3 3 3 4 3 3 4 3 3 3 3 One vertical, one horizontal.   Same level.   Same column. 14   14 N Top Bottom Right side Left side 3 2 3 3 8 6 7 7 4 3 3 3 4 3 4 4 4 3 3 3 Both horizontal.   Same level.   Opposite columns. 15   15 N Top Bottom Right side Left side 3 3 3 3 8 7 7 8 4 3 3 4 4 4 4 4 4 3 3 4 One vertical, one horizontal.   Different levels.   Opposite columns. 16   16 O Top Bottom Right side Left side 3 4 3 3 8 8 8 8 4 3 4 4 4 4 4 4 4 4 4 4 Both vertical.   Same level.   Opposite columns. 17   17 O Top Bottom Right side Left side 3 4 4 3 8 9 9 8 4 4 4 4 4 4 4 4 4 5 5 4 One vertical, one horizontal.   Same level.   Same column. 18   18 O Top Bottom Right side Left side 3 4 4 4 8 10 9 9 4 5 4 4 4 5 4 4 4 5 5 5 Both horizontal.   Same level.   Opposite columns. 19   19 F Top Bottom Right side Left side 4 4 4 4 9 10 10 9 4 5 5 4 4 5 4 4 5 5 5 5 One vertical, one horizontal.   Different levels.   Opposite columns. 20   20 Ne Top Bottom Right side Left side 5 4 4 4 10 10 10 10 4 5 5 5 5 5 5 5 5 5 5 5 Both vertical.   Same level.   Opposite columns. 21   21 Ne Top Bottom Right side Left side 5 4 4 5 11 10 10 11 5 5 5 5 6 5 56 5 5 5 5 One vertical, one horizontal.   Same level.   Same column. 22   22 Ne Top Bottom Right side Left side 5 4 5 5 12 10 11 11 6 5 5 5 6 5 6 6 6 5 5 5 Both horizontal.   Same level.   Opposite columns. 23   Na Top Bottom Right side Left side 5 5 5 5 12 11 11 12 6 5 5 6 6 6 6 6 6 5 5 6 One vertical, one horizontal.   Different levels.   Opposite columns. 24   24 Mg Top Bottom Right side Left side 5 6 5 5 12 12 12 12 6 5 6 6 6 6 6 6 6 6 6 6 Both vertical.   Same level.   Opposite columns. 25   25 Mg Top Bottom Right side Left side 5 6 6 5 12 13 13 12 6 6 6 6 6 6 6 6 6 7 7 6 One vertical, one horizontal.   Same level.   Same column. 26   26 Mg Top Bottom Right side Left side 5 6 6 6 12 14 13 13 6 7 6 6 6 7 6 6 6 7 7 7 Both horizontal.   Same level.   Opposite columns. 27   Al Top Bottom Right side Left side 6 6 6 6 13 14 14 13 6 7 7 6 6 7 7 6 7 7 7 7 One vertical, one horizontal.   Different levels.   Opposite columns. 28   28 Si Top Bottom Right side Left side 7 6 6 6 14 14 14 14 6 7 7 7 7 7 7 7 7 7 7 7 Both vertical.   Same level.   Opposite columns. 29   29 Si Top Bottom Right side Left side 7 6 6 7 15 14 14 15 7 7 7 7 8 7 7 8 7 7 7 7 One vertical, one horizontal.   Same level.   Same column. 30   30 Si Top Bottom Right side Left side 7 6 7 7 16 14 15 15 8 7 7 7 8 7 8 8 8 7 7 7 Both horizontal.   Same level.   Opposite columns. 31   P Top Bottom Right side Left side 7 7 7 7 16 15 15 16 8 7 7 8 8 8 8 8 8 7 7 8 One vertical, one horizontal.   Different levels.   Opposite columns. 32   32 S Top Bottom Right side Left side 7 8 7 7 1616 16 16 8 7 8 8 8 8 8 8 8 8 8 8 Both vertical.   Same level.   Opposite columns. 33   33 S Top Bottom Right side Left side 7 8 8 7 16 1717 16 8 8 8 8 8 8 8 8 8 9 9 8 One vertical, one horizontal.   Same level.   Same column. 34   34 S Top Bottom Right side Left side 7 8 8 8 16 18 17 17 8 9 8 8 8 9 8 8 8 9 9 9 Both horizontal.   Same level.   Opposite columns. 35   35 Cl Top Bottom Right side Left side 8 8 8 8 17 18 18 17 8 9 9 8 8 9 9 8 9 9 9 9 One vertical, one horizontal.   Different levels.   Opposite columns. 36   36 Ar Top Bottom Right side Left side 9 8 8 8 18 1818 18 8 9 9 9 9 9 9 9 9 9 9 9 Both vertical.   Same level.   Opposite columns.

This chart can be tentatively assigned to all isotopes up to Ar36.   From there starting with the pair Ar36 – S36 there are pairs of isotopes with the same atomic mass. These pairs cannot be explained from the sequence of adding a nucleon on either end, as either addition forms only one nucleus of the same arrangement, so the properties would be the same. So how are, for example, Ar36 and S36 to be distinguished?

Length to Width Ratio of the Primary Particle Rods

Consider this that adding a nucleon to a side gives a point of branching to the shape and properties of a nucleus of equal mass.   Perhaps a nucleon gets added at C135 to the side because the width of the C135 nucleus now equals the width of a rod, so this side nucleon is “shielded” from the PP flow that would blow it off if it accreted when it had an overlap on either side of the nucleus.   This might provide a basis then for determining the length to width ratio of a rod. Each nucleon is 4 rod widths wide (2 faces and 2 extras) but as any 2 nucleons share an extra, each nucleon added to Hydrogen increases the width of the nucleus by 3.   The Hydrogen width alone = 4.   For 2 nucleons the total width = 7 (rod widths) and for 3 nucleons, total width = 10, etc., or:

eq. 5-1           atomic mass x 3 + 1 = width of respective nucleus

So for C135,   35 x 3 = 105 + 1 = 106 length/width ratio, or a value near such. For the fine structure constant to equal the rod width the side accretion would need to begin around C46.   Looking at the Chart of the Nuclides [1] , branching, as said above, starts at Ar36 – S36.   Therefore either the fine structure constant is not equal to a rod width:length ratio or side accretion starts at a point other than full rod width.

Seventy-five percent of 137 (“the fine structure constant”) equals 102.75.   This amount is first satisfied at S34 (34 X 3 + 1 = 103).   This would leave a 25% overhang, considering the way a nucleon would rotate on PP’s hitting it, 75% contact with the nucleus may be a very auspicious amount for all such interactions to be diverted back into the nucleus and uniform motions maintained. Because then it might be possible to have the width to length ratio of a rod within the range for the fine structure constant; 1 : 137.036…, and this would tie in well with standard physics, this will be the first assumption of what happens.

This would leave the nucleon then as composed of 137 + 137 + 2 (extras) or 276 (identical) primary particles. But I consider the last two on each face too tight a fit, so 135 + 135 + 2 or 272 is more likely.

Now how the elements over S34 are figured is a matter I can’t explain, but some tentative ideas make it seem somewhat feasible. Assuming that the nucleus always has a certain permanent spin running around the sides, due to the first accretion of H + H = deuterium having the extras in the manner for this spin (see figure 5.1 and associated discussion).   Therefore all accretion must be on the right or left side.   From Chart 5.1, we see that at S34 each side has an unequal number of points of contact in the a & d rows.   At C135 however the right side has 9 contact points in row a and 9 in row d, therefore a nucleon can, and does, get added. This is diagrammed in Figure 5.5 (the center of the mass of the nucleon falls on the middle contact point leaving four balanced contact points on either side).

Figure 5.5
Top view of the nucleus up to Ar36 and beyond.

KEY        Parts of the length are cut out, as represented.

Each rectangle represents a nucleon. Dark rectangles are “lower level” nucleons; light rectangles are “upper level” nucleons.

#35 & #36 are needed to balance the contact points on respective sides, with side nucleons centered into a gap.

C135 + a side accretion then becomes S36.   C135 + an end accretion still becomes Ar36.  Likewise, Ar36 having no side, if it accreted an end next, would be expected to be Ar37 and stable, which doesn’t occur. Therefore, assume Ar36 cannot accrete an end, only a side, and that must become CL37.

Now from FIG. 5.5 we see that for both sides to be added, spots #35 & #36 must be filled.  At S36 spot #36 is empty, so an accretion can occur at #36, #37, or the side

with a nucleon already.   Now, if spot #36 is filled the accretion matches the CL37 from Ar36 and this seems likely so assume such.   This is diagrammed in Figure 5.6.From here many sequences might be supposed (see also chapter 6, electron orbitals).

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