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.

 

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.

   

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

       

           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. With one full face and backed up by a half face, it would be suffices to have the extras perpendicular to each other, but this scenario would seem to only produce spinning N, at least over the long run

         T he extras would need to be at center of mass in order for 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. 

 

           I now conclude that for a P falling onto the face of a N it would produce rotation (spin) of the N.

            Only the hits achieving balance on the extras would directly produce linear motion of a N.

            But, I have thought out but need to re-figure, on most or some additional P impacts to N, when spin is stopped a sequence of motion within the P of the N, it ends with an extra being rotated back into N producing linear motion again. So, this needs to definitely get that figured out, then meld all this back into the chapters.

           Okay here is an example of this. A P overtakes a N hitting on the face. Unison motion is established, the rest goes into torque, (other considerations may need to be considered here too) then falls across face of N this causes rotation of P which gives rotation to face at points of contact. This causes those rods to rotate on the P which cause their backswing to hit the other face. This causes re-rotation of the first face into the other face but only those rods that had been in contact with the P, so the far face rotates on those rods causing all of them to rotate into the first face which also then rotates the extra on that side and the P, as well as the back extra being swept up in this motion. So, the whole N and the P are in a rotation in addition to their linear motion.

           When another P (Px) hits, say a direct hit on the extra, all motion goes through the extra into torque, then re-rotation into the face and the rest of the N as linear motion.

          BUT with the P across face still it would cause all to collapse to rotational motion again, now with no linear motion. Here I see no way to get the P off or produce linear, except perhaps after a second layer of plastered P’s is formed it might rotate into the extra which would re-rotate and accelerate the N linear, leaving the plastered P’s in its wake. How the to get an accelerated P as a fast electron is another story. Very complicated here, and with winter here and need to keep woodstove loaded with wood and feed the animals may not have time or energy to do much.

 

Outlines help!

PP off N          Direct Hits   also   overtaking hits

1.        across face (after torque and rotation)

a)      Adjacent face

b)      Far face

2.      Off extra

A)     Rotates to midpoint and accelerates

B)      Rotates to edge

Then question of two of these things happening at once

The Second Layer overtaking and direct

1.      Across extra and plastered P (likely)  near and far faces (IF ROTATES AROUND TO FRONT)

 





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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