A Non Fiction Trilogy

CHAPTER 5 ELEMENTS

Jan 22, 2019 changes needed in line with chapter 4 at note at top of page and as below

June 9,2018

This will need changes as chapters 3-4.5 have been. It should be more in line with the thoughts expressed in red below and those in the last chapter.

Spinning and linear should be able to be combined I believe, it’s a complicated and dynamic situation where if linear is already present, spin can be added on. Else if any hit (overtaking hit) occurs spin must stop, and then the resulting motion would turn spin speed into a linear acceleration (speed) of the Nucleon(s). So spinning and linear for one N okay, but I may still have problem for two if linked as I used to have it.

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 must be, on each face, running in opposite directions.

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).

[1]               Put out by General Electric

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